Calculus Examples

$\int \frac{x}{16 x^{4} - 1} d x$

Step 1

Write the fraction using partial fraction decomposition.

Tap for more steps...

Step 1.1

Decompose the fraction and multiply through by the common denominator.

Tap for more steps...

Step 1.1.1

Factor the fraction.

Tap for more steps...

Step 1.1.1.1

Rewrite $16 x^{4}$ as ${(4 x^{2})}^{2}$ .

$\frac{x}{{(4 x^{2})}^{2} - 1}$

Step 1.1.1.2

Rewrite $1$ as $1^{2}$ .

$\frac{x}{{(4 x^{2})}^{2} - 1^{2}}$

Step 1.1.1.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4 x^{2}$ and $b = 1$ .

$\frac{x}{(4 x^{2} + 1) (4 x^{2} - 1)}$

Step 1.1.1.4

Simplify.

Tap for more steps...

Step 1.1.1.4.1

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$\frac{x}{(4 x^{2} + 1) ({(2 x)}^{2} - 1)}$

Step 1.1.1.4.2

Rewrite $1$ as $1^{2}$ .

$\frac{x}{(4 x^{2} + 1) ({(2 x)}^{2} - 1^{2})}$

Step 1.1.1.4.3

Factor.

Tap for more steps...

Step 1.1.1.4.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 1$ .

$\frac{x}{(4 x^{2} + 1) ((2 x + 1) (2 x - 1))}$

Step 1.1.1.4.3.2

Remove unnecessary parentheses.

$\frac{x}{(4 x^{2} + 1) (2 x + 1) (2 x - 1)}$

Step 1.1.2

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor is 2nd order, $2$ terms are required in the numerator. The number of terms required in the numerator is always equal to the order of the factor in the denominator.

$\frac{A x + B}{4 x^{2} + 1}$

Step 1.1.3

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $C$ .

$\frac{A x + B}{4 x^{2} + 1} + \frac{C}{2 x + 1}$

Step 1.1.4

$\frac{A x + B}{4 x^{2} + 1} + \frac{C}{2 x + 1} + \frac{D}{2 x - 1}$

Step 1.1.5

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(4 x^{2} + 1) (2 x + 1) (2 x - 1)$ .

$\frac{x (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{(4 x^{2} + 1) (2 x + 1) (2 x - 1)} = \frac{(A x + B) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{4 x^{2} + 1} + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.6

Cancel the common factor of $4 x^{2} + 1$ .

Tap for more steps...

Step 1.1.6.1

Cancel the common factor.

Step 1.1.6.2

Rewrite the expression.

$\frac{(x (2 x + 1)) (2 x - 1)}{(2 x + 1) (2 x - 1)} = \frac{(A x + B) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{4 x^{2} + 1} + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.7

Cancel the common factor of $2 x + 1$ .

Tap for more steps...

Step 1.1.7.1

Cancel the common factor.

$\frac{x (2 x + 1) (2 x - 1)}{(2 x + 1) (2 x - 1)} = \frac{(A x + B) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{4 x^{2} + 1} + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.7.2

Rewrite the expression.

$\frac{(x) (2 x - 1)}{2 x - 1} = \frac{(A x + B) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{4 x^{2} + 1} + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.8

Cancel the common factor of $2 x - 1$ .

Tap for more steps...

Step 1.1.8.1

Cancel the common factor.

$\frac{x (2 x - 1)}{2 x - 1} = \frac{(A x + B) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{4 x^{2} + 1} + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.8.2

Divide $x$ by $1$ .

$x = \frac{(A x + B) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{4 x^{2} + 1} + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9

Simplify each term.

Tap for more steps...

Step 1.1.9.1

Cancel the common factor of $4 x^{2} + 1$ .

Tap for more steps...

Step 1.1.9.1.1

Cancel the common factor.

$x = \frac{(A x + B) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{4 x^{2} + 1} + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.1.2

Divide $((A x + B) (2 x + 1)) (2 x - 1)$ by $1$ .

$x = ((A x + B) (2 x + 1)) (2 x - 1) + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.2

Expand $(A x + B) (2 x + 1)$ using the FOIL Method.

Tap for more steps...

Step 1.1.9.2.1

Apply the distributive property.

$x = (A x (2 x + 1) + B (2 x + 1)) (2 x - 1) + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.2.2

Apply the distributive property.

$x = (A x (2 x) + A x \cdot 1 + B (2 x + 1)) (2 x - 1) + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.2.3

Apply the distributive property.

$x = (A x (2 x) + A x \cdot 1 + B (2 x) + B \cdot 1) (2 x - 1) + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.3

Simplify each term.

Tap for more steps...

Step 1.1.9.3.1

Rewrite using the commutative property of multiplication.

$x = (2 A x \cdot x + A x \cdot 1 + B (2 x) + B \cdot 1) (2 x - 1) + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.3.2

Multiply $x$ by $x$ by adding the exponents.

Tap for more steps...

Step 1.1.9.3.2.1

Move $x$ .

$x = (2 A (x \cdot x) + A x \cdot 1 + B (2 x) + B \cdot 1) (2 x - 1) + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.3.2.2

Multiply $x$ by $x$ .

$x = (2 A x^{2} + A x \cdot 1 + B (2 x) + B \cdot 1) (2 x - 1) + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.3.3

Multiply $A$ by $1$ .

$x = (2 A x^{2} + A x + B (2 x) + B \cdot 1) (2 x - 1) + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.3.4

Rewrite using the commutative property of multiplication.

$x = (2 A x^{2} + A x + 2 B x + B \cdot 1) (2 x - 1) + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.3.5

Multiply $B$ by $1$ .

$x = (2 A x^{2} + A x + 2 B x + B) (2 x - 1) + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.4

Expand $(2 A x^{2} + A x + 2 B x + B) (2 x - 1)$ by multiplying each term in the first expression by each term in the second expression.

$x = 2 A x^{2} (2 x) + 2 A x^{2} \cdot - 1 + A x (2 x) + A x \cdot - 1 + 2 B x (2 x) + 2 B x \cdot - 1 + B (2 x) + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5

Simplify each term.

Tap for more steps...

Step 1.1.9.5.1

Multiply $x^{2}$ by $x$ by adding the exponents.

Tap for more steps...

Step 1.1.9.5.1.1

Move $x$ .

$x = 2 A (x \cdot x^{2}) \cdot 2 + 2 A x^{2} \cdot - 1 + A x (2 x) + A x \cdot - 1 + 2 B x (2 x) + 2 B x \cdot - 1 + B (2 x) + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.1.2

Multiply $x$ by $x^{2}$ .

Tap for more steps...

Step 1.1.9.5.1.2.1

Raise $x$ to the power of $1$ .

Step 1.1.9.5.1.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x = 2 A x^{1 + 2} \cdot 2 + 2 A x^{2} \cdot - 1 + A x (2 x) + A x \cdot - 1 + 2 B x (2 x) + 2 B x \cdot - 1 + B (2 x) + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.1.3

Add $1$ and $2$ .

$x = 2 A x^{3} \cdot 2 + 2 A x^{2} \cdot - 1 + A x (2 x) + A x \cdot - 1 + 2 B x (2 x) + 2 B x \cdot - 1 + B (2 x) + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.2

Multiply $2$ by $2$ .

$x = 4 A x^{3} + 2 A x^{2} \cdot - 1 + A x (2 x) + A x \cdot - 1 + 2 B x (2 x) + 2 B x \cdot - 1 + B (2 x) + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.3

Multiply $- 1$ by $2$ .

$x = 4 A x^{3} - 2 A x^{2} + A x (2 x) + A x \cdot - 1 + 2 B x (2 x) + 2 B x \cdot - 1 + B (2 x) + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.4

Rewrite using the commutative property of multiplication.

$x = 4 A x^{3} - 2 A x^{2} + 2 A x \cdot x + A x \cdot - 1 + 2 B x (2 x) + 2 B x \cdot - 1 + B (2 x) + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.5

Multiply $x$ by $x$ by adding the exponents.

Tap for more steps...

Step 1.1.9.5.5.1

Move $x$ .

$x = 4 A x^{3} - 2 A x^{2} + 2 A (x \cdot x) + A x \cdot - 1 + 2 B x (2 x) + 2 B x \cdot - 1 + B (2 x) + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.5.2

Multiply $x$ by $x$ .

$x = 4 A x^{3} - 2 A x^{2} + 2 A x^{2} + A x \cdot - 1 + 2 B x (2 x) + 2 B x \cdot - 1 + B (2 x) + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.6

Move $- 1$ to the left of $A x$ .

$x = 4 A x^{3} - 2 A x^{2} + 2 A x^{2} - 1 \cdot (A x) + 2 B x (2 x) + 2 B x \cdot - 1 + B (2 x) + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.7

Rewrite $- 1 (A x)$ as $- (A x)$ .

$x = 4 A x^{3} - 2 A x^{2} + 2 A x^{2} - (A x) + 2 B x (2 x) + 2 B x \cdot - 1 + B (2 x) + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.8

Multiply $x$ by $x$ by adding the exponents.

Tap for more steps...

Step 1.1.9.5.8.1

Move $x$ .

$x = 4 A x^{3} - 2 A x^{2} + 2 A x^{2} - A x + 2 B (x \cdot x) \cdot 2 + 2 B x \cdot - 1 + B (2 x) + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.8.2

Multiply $x$ by $x$ .

$x = 4 A x^{3} - 2 A x^{2} + 2 A x^{2} - A x + 2 B x^{2} \cdot 2 + 2 B x \cdot - 1 + B (2 x) + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.9

Multiply $2$ by $2$ .

$x = 4 A x^{3} - 2 A x^{2} + 2 A x^{2} - A x + 4 B x^{2} + 2 B x \cdot - 1 + B (2 x) + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.10

Multiply $- 1$ by $2$ .

$x = 4 A x^{3} - 2 A x^{2} + 2 A x^{2} - A x + 4 B x^{2} - 2 B x + B (2 x) + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.11

Rewrite using the commutative property of multiplication.

$x = 4 A x^{3} - 2 A x^{2} + 2 A x^{2} - A x + 4 B x^{2} - 2 B x + 2 B x + B \cdot - 1 + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.12

Move $- 1$ to the left of $B$ .

$x = 4 A x^{3} - 2 A x^{2} + 2 A x^{2} - A x + 4 B x^{2} - 2 B x + 2 B x - 1 \cdot B + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.5.13

Rewrite $- 1 B$ as $- B$ .

$x = 4 A x^{3} - 2 A x^{2} + 2 A x^{2} - A x + 4 B x^{2} - 2 B x + 2 B x - B + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.6

Combine the opposite terms in $4 A x^{3} - 2 A x^{2} + 2 A x^{2} - A x + 4 B x^{2} - 2 B x + 2 B x - B$ .

Tap for more steps...

Step 1.1.9.6.1

Add $- 2 A x^{2}$ and $2 A x^{2}$ .

$x = 4 A x^{3} + 0 - A x + 4 B x^{2} - 2 B x + 2 B x - B + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.6.2

Add $4 A x^{3}$ and $0$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - 2 B x + 2 B x - B + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.6.3

Add $- 2 B x$ and $2 B x$ .

$x = 4 A x^{3} - A x + 4 B x^{2} + 0 - B + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.6.4

Add $4 A x^{3} - A x + 4 B x^{2}$ and $0$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.7

Cancel the common factor of $2 x + 1$ .

Tap for more steps...

Step 1.1.9.7.1

Cancel the common factor.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + \frac{(C) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x + 1} + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.7.2

Divide $(C) (4 x^{2} + 1) (2 x - 1)$ by $1$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + (C) (4 x^{2} + 1) (2 x - 1) + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.8

Apply the distributive property.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + (C (4 x^{2}) + C \cdot 1) (2 x - 1) + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.9

Rewrite using the commutative property of multiplication.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + (4 C x^{2} + C \cdot 1) (2 x - 1) + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.10

Multiply $C$ by $1$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + (4 C x^{2} + C) (2 x - 1) + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.11

Expand $(4 C x^{2} + C) (2 x - 1)$ using the FOIL Method.

Tap for more steps...

Step 1.1.9.11.1

Apply the distributive property.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 4 C x^{2} (2 x - 1) + C (2 x - 1) + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.11.2

Apply the distributive property.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 4 C x^{2} (2 x) + 4 C x^{2} \cdot - 1 + C (2 x - 1) + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.11.3

Apply the distributive property.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 4 C x^{2} (2 x) + 4 C x^{2} \cdot - 1 + C (2 x) + C \cdot - 1 + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.12

Simplify each term.

Tap for more steps...

Step 1.1.9.12.1

Multiply $x^{2}$ by $x$ by adding the exponents.

Tap for more steps...

Step 1.1.9.12.1.1

Move $x$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 4 C (x \cdot x^{2}) \cdot 2 + 4 C x^{2} \cdot - 1 + C (2 x) + C \cdot - 1 + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.12.1.2

Multiply $x$ by $x^{2}$ .

Tap for more steps...

Step 1.1.9.12.1.2.1

Raise $x$ to the power of $1$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 4 C (x \cdot x^{2}) \cdot 2 + 4 C x^{2} \cdot - 1 + C (2 x) + C \cdot - 1 + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.12.1.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 4 C x^{1 + 2} \cdot 2 + 4 C x^{2} \cdot - 1 + C (2 x) + C \cdot - 1 + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.12.1.3

Add $1$ and $2$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 4 C x^{3} \cdot 2 + 4 C x^{2} \cdot - 1 + C (2 x) + C \cdot - 1 + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.12.2

Multiply $2$ by $4$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} + 4 C x^{2} \cdot - 1 + C (2 x) + C \cdot - 1 + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.12.3

Multiply $- 1$ by $4$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + C (2 x) + C \cdot - 1 + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.12.4

Rewrite using the commutative property of multiplication.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x + C \cdot - 1 + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.12.5

Move $- 1$ to the left of $C$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - 1 \cdot C + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.12.6

Rewrite $- 1 C$ as $- C$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.13

Cancel the common factor of $2 x - 1$ .

Tap for more steps...

Step 1.1.9.13.1

Cancel the common factor.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + \frac{(D) (4 x^{2} + 1) (2 x + 1) (2 x - 1)}{2 x - 1}$

Step 1.1.9.13.2

Divide $(D) (4 x^{2} + 1) (2 x + 1)$ by $1$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + (D) (4 x^{2} + 1) (2 x + 1)$

Step 1.1.9.14

Apply the distributive property.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + (D (4 x^{2}) + D \cdot 1) (2 x + 1)$

Step 1.1.9.15

Rewrite using the commutative property of multiplication.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + (4 D x^{2} + D \cdot 1) (2 x + 1)$

Step 1.1.9.16

Multiply $D$ by $1$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + (4 D x^{2} + D) (2 x + 1)$

Step 1.1.9.17

Expand $(4 D x^{2} + D) (2 x + 1)$ using the FOIL Method.

Tap for more steps...

Step 1.1.9.17.1

Apply the distributive property.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + 4 D x^{2} (2 x + 1) + D (2 x + 1)$

Step 1.1.9.17.2

Apply the distributive property.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + 4 D x^{2} (2 x) + 4 D x^{2} \cdot 1 + D (2 x + 1)$

Step 1.1.9.17.3

Apply the distributive property.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + 4 D x^{2} (2 x) + 4 D x^{2} \cdot 1 + D (2 x) + D \cdot 1$

Step 1.1.9.18

Simplify each term.

Tap for more steps...

Step 1.1.9.18.1

Multiply $x^{2}$ by $x$ by adding the exponents.

Tap for more steps...

Step 1.1.9.18.1.1

Move $x$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + 4 D (x \cdot x^{2}) \cdot 2 + 4 D x^{2} \cdot 1 + D (2 x) + D \cdot 1$

Step 1.1.9.18.1.2

Multiply $x$ by $x^{2}$ .

Tap for more steps...

Step 1.1.9.18.1.2.1

Raise $x$ to the power of $1$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + 4 D (x \cdot x^{2}) \cdot 2 + 4 D x^{2} \cdot 1 + D (2 x) + D \cdot 1$

Step 1.1.9.18.1.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + 4 D x^{1 + 2} \cdot 2 + 4 D x^{2} \cdot 1 + D (2 x) + D \cdot 1$

Step 1.1.9.18.1.3

Add $1$ and $2$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + 4 D x^{3} \cdot 2 + 4 D x^{2} \cdot 1 + D (2 x) + D \cdot 1$

Step 1.1.9.18.2

Multiply $2$ by $4$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + 8 D x^{3} + 4 D x^{2} \cdot 1 + D (2 x) + D \cdot 1$

Step 1.1.9.18.3

Multiply $4$ by $1$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + 8 D x^{3} + 4 D x^{2} + D (2 x) + D \cdot 1$

Step 1.1.9.18.4

Rewrite using the commutative property of multiplication.

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + 8 D x^{3} + 4 D x^{2} + 2 D x + D \cdot 1$

Step 1.1.9.18.5

Multiply $D$ by $1$ .

$x = 4 A x^{3} - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + 8 D x^{3} + 4 D x^{2} + 2 D x + D$

Step 1.1.10

Reorder.

Tap for more steps...

Step 1.1.10.1

Move $A$ .

$x = 4 x^{3} A - A x + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + 8 D x^{3} + 4 D x^{2} + 2 D x + D$

Step 1.1.10.2

Move $A$ .

$x = 4 x^{3} A - 1 x A + 4 B x^{2} - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + 8 D x^{3} + 4 D x^{2} + 2 D x + D$

Step 1.1.10.3

Move $B$ .

$x = 4 x^{3} A - 1 x A + 4 x^{2} B - B + 8 C x^{3} - 4 C x^{2} + 2 C x - C + 8 D x^{3} + 4 D x^{2} + 2 D x + D$

Step 1.1.10.4

Move $- C$ .

$x = 4 x^{3} A - 1 x A + 4 x^{2} B - B + 8 C x^{3} - 4 C x^{2} + 2 C x + 8 D x^{3} + 4 D x^{2} + 2 D x - C + D$

Step 1.1.10.5

Move $- B$ .

$x = 4 x^{3} A - 1 x A + 4 x^{2} B + 8 C x^{3} - 4 C x^{2} + 2 C x + 8 D x^{3} + 4 D x^{2} + 2 D x - B - C + D$

Step 1.1.10.6

Move $2 C x$ .

$x = 4 x^{3} A - 1 x A + 4 x^{2} B + 8 C x^{3} - 4 C x^{2} + 8 D x^{3} + 4 D x^{2} + 2 C x + 2 D x - B - C + D$

Step 1.1.10.7

Move $- 1 x A$ .

$x = 4 x^{3} A + 4 x^{2} B + 8 C x^{3} - 4 C x^{2} + 8 D x^{3} + 4 D x^{2} - 1 x A + 2 C x + 2 D x - B - C + D$

Step 1.1.10.8

Move $- 4 C x^{2}$ .

$x = 4 x^{3} A + 4 x^{2} B + 8 C x^{3} + 8 D x^{3} - 4 C x^{2} + 4 D x^{2} - 1 x A + 2 C x + 2 D x - B - C + D$

Step 1.1.10.9

Move $4 x^{2} B$ .

$x = 4 x^{3} A + 8 C x^{3} + 8 D x^{3} + 4 x^{2} B - 4 C x^{2} + 4 D x^{2} - 1 x A + 2 C x + 2 D x - B - C + D$

Step 1.2

Create equations for the partial fraction variables and use them to set up a system of equations.

Tap for more steps...

Step 1.2.1

Create an equation for the partial fraction variables by equating the coefficients of $x^{3}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = 4 A + 8 C + 8 D$

Step 1.2.2

Create an equation for the partial fraction variables by equating the coefficients of $x^{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = 4 B - 4 C + 4 D$

Step 1.2.3

Create an equation for the partial fraction variables by equating the coefficients of $x$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = - 1 A + 2 C + 2 D$

Step 1.2.4

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $x$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = - 1 B - 1 C + D$

Step 1.2.5

Set up the system of equations to find the coefficients of the partial fractions.

$0 = 4 A + 8 C + 8 D$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

$0 = 4 A + 8 C + 8 D$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3

Solve the system of equations.

Tap for more steps...

Step 1.3.1

Solve for $A$ in $0 = 4 A + 8 C + 8 D$ .

Tap for more steps...

Step 1.3.1.1

Rewrite the equation as $4 A + 8 C + 8 D = 0$ .

$4 A + 8 C + 8 D = 0$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.1.2

Move all terms not containing $A$ to the right side of the equation.

Tap for more steps...

Step 1.3.1.2.1

Subtract $8 C$ from both sides of the equation.

$4 A + 8 D = - 8 C$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.1.2.2

Subtract $8 D$ from both sides of the equation.

$4 A = - 8 C - 8 D$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

$4 A = - 8 C - 8 D$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.1.3

Divide each term in $4 A = - 8 C - 8 D$ by $4$ and simplify.

Tap for more steps...

Step 1.3.1.3.1

Divide each term in $4 A = - 8 C - 8 D$ by $4$ .

$\frac{4 A}{4} = \frac{- 8 C}{4} + \frac{- 8 D}{4}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.1.3.2

Simplify the left side.

Tap for more steps...

Step 1.3.1.3.2.1

Cancel the common factor of $4$ .

Tap for more steps...

Step 1.3.1.3.2.1.1

Cancel the common factor.

$\frac{4 A}{4} = \frac{- 8 C}{4} + \frac{- 8 D}{4}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.1.3.2.1.2

Divide $A$ by $1$ .

$A = \frac{- 8 C}{4} + \frac{- 8 D}{4}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

$A = \frac{- 8 C}{4} + \frac{- 8 D}{4}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

$A = \frac{- 8 C}{4} + \frac{- 8 D}{4}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.1.3.3

Simplify the right side.

Tap for more steps...

Step 1.3.1.3.3.1

Simplify each term.

Tap for more steps...

Step 1.3.1.3.3.1.1

Cancel the common factor of $- 8$ and $4$ .

Tap for more steps...

Step 1.3.1.3.3.1.1.1

Factor $4$ out of $- 8 C$ .

$A = \frac{4 (- 2 C)}{4} + \frac{- 8 D}{4}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.1.3.3.1.1.2

Cancel the common factors.

Tap for more steps...

Step 1.3.1.3.3.1.1.2.1

Factor $4$ out of $4$ .

$A = \frac{4 (- 2 C)}{4 (1)} + \frac{- 8 D}{4}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.1.3.3.1.1.2.2

Cancel the common factor.

$A = \frac{4 (- 2 C)}{4 \cdot 1} + \frac{- 8 D}{4}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.1.3.3.1.1.2.3

Rewrite the expression.

$A = \frac{- 2 C}{1} + \frac{- 8 D}{4}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.1.3.3.1.1.2.4

Divide $- 2 C$ by $1$ .

$A = - 2 C + \frac{- 8 D}{4}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

$A = - 2 C + \frac{- 8 D}{4}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

$A = - 2 C + \frac{- 8 D}{4}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.1.3.3.1.2

Cancel the common factor of $- 8$ and $4$ .

Tap for more steps...

Step 1.3.1.3.3.1.2.1

Factor $4$ out of $- 8 D$ .

$A = - 2 C + \frac{4 (- 2 D)}{4}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.1.3.3.1.2.2

Cancel the common factors.

Tap for more steps...

Step 1.3.1.3.3.1.2.2.1

Factor $4$ out of $4$ .

$A = - 2 C + \frac{4 (- 2 D)}{4 (1)}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.1.3.3.1.2.2.2

Cancel the common factor.

$A = - 2 C + \frac{4 (- 2 D)}{4 \cdot 1}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.1.3.3.1.2.2.3

Rewrite the expression.

$A = - 2 C + \frac{- 2 D}{1}$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.1.3.3.1.2.2.4

Divide $- 2 D$ by $1$ .

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$1 = - 1 A + 2 C + 2 D$

$0 = - 1 B - 1 C + D$

Step 1.3.2

Replace all occurrences of $A$ with $- 2 C - 2 D$ in each equation.

Tap for more steps...

Step 1.3.2.1

Replace all occurrences of $A$ in $1 = - 1 A + 2 C + 2 D$ with $- 2 C - 2 D$ .

$1 = - 1 (- 2 C - 2 D) + 2 C + 2 D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.2.2

Simplify the right side.

Tap for more steps...

Step 1.3.2.2.1

Simplify $- 1 (- 2 C - 2 D) + 2 C + 2 D$ .

Tap for more steps...

Step 1.3.2.2.1.1

Simplify each term.

Tap for more steps...

Step 1.3.2.2.1.1.1

Apply the distributive property.

$1 = - 1 (- 2 C) - 1 (- 2 D) + 2 C + 2 D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.2.2.1.1.2

Multiply $- 2$ by $- 1$ .

$1 = 2 C - 1 (- 2 D) + 2 C + 2 D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.2.2.1.1.3

Multiply $- 2$ by $- 1$ .

$1 = 2 C + 2 D + 2 C + 2 D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$1 = 2 C + 2 D + 2 C + 2 D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.2.2.1.2

Simplify by adding terms.

Tap for more steps...

Step 1.3.2.2.1.2.1

Add $2 C$ and $2 C$ .

$1 = 4 C + 2 D + 2 D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.2.2.1.2.2

Add $2 D$ and $2 D$ .

$1 = 4 C + 4 D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$1 = 4 C + 4 D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$1 = 4 C + 4 D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$1 = 4 C + 4 D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$1 = 4 C + 4 D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.3

Solve for $C$ in $1 = 4 C + 4 D$ .

Tap for more steps...

Step 1.3.3.1

Rewrite the equation as $4 C + 4 D = 1$ .

$4 C + 4 D = 1$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.3.2

Subtract $4 D$ from both sides of the equation.

$4 C = 1 - 4 D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.3.3

Divide each term in $4 C = 1 - 4 D$ by $4$ and simplify.

Tap for more steps...

Step 1.3.3.3.1

Divide each term in $4 C = 1 - 4 D$ by $4$ .

$\frac{4 C}{4} = \frac{1}{4} + \frac{- 4 D}{4}$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.3.3.2

Simplify the left side.

Tap for more steps...

Step 1.3.3.3.2.1

Cancel the common factor of $4$ .

Tap for more steps...

Step 1.3.3.3.2.1.1

Cancel the common factor.

$\frac{4 C}{4} = \frac{1}{4} + \frac{- 4 D}{4}$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.3.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{1}{4} + \frac{- 4 D}{4}$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$C = \frac{1}{4} + \frac{- 4 D}{4}$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$C = \frac{1}{4} + \frac{- 4 D}{4}$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.3.3.3

Simplify the right side.

Tap for more steps...

Step 1.3.3.3.3.1

Cancel the common factor of $- 4$ and $4$ .

Tap for more steps...

Step 1.3.3.3.3.1.1

Factor $4$ out of $- 4 D$ .

$C = \frac{1}{4} + \frac{4 (- D)}{4}$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.3.3.3.1.2

Cancel the common factors.

Tap for more steps...

Step 1.3.3.3.3.1.2.1

Factor $4$ out of $4$ .

$C = \frac{1}{4} + \frac{4 (- D)}{4 (1)}$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.3.3.3.1.2.2

Cancel the common factor.

$C = \frac{1}{4} + \frac{4 (- D)}{4 \cdot 1}$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.3.3.3.1.2.3

Rewrite the expression.

$C = \frac{1}{4} + \frac{- D}{1}$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.3.3.3.1.2.4

Divide $- D$ by $1$ .

$C = \frac{1}{4} - D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$C = \frac{1}{4} - D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$C = \frac{1}{4} - D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$C = \frac{1}{4} - D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$C = \frac{1}{4} - D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$C = \frac{1}{4} - D$

$A = - 2 C - 2 D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.4

Replace all occurrences of $C$ with $\frac{1}{4} - D$ in each equation.

Tap for more steps...

Step 1.3.4.1

Replace all occurrences of $C$ in $A = - 2 C - 2 D$ with $\frac{1}{4} - D$ .

$A = - 2 (\frac{1}{4} - D) - 2 D$

$C = \frac{1}{4} - D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.2

Simplify the right side.

Tap for more steps...

Step 1.3.4.2.1

Simplify $- 2 (\frac{1}{4} - D) - 2 D$ .

Tap for more steps...

Step 1.3.4.2.1.1

Simplify each term.

Tap for more steps...

Step 1.3.4.2.1.1.1

Apply the distributive property.

$A = - 2 (\frac{1}{4}) - 2 (- D) - 2 D$

$C = \frac{1}{4} - D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.2.1.1.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.3.4.2.1.1.2.1

Factor $2$ out of $- 2$ .

$A = 2 (- 1) (\frac{1}{4}) - 2 (- D) - 2 D$

$C = \frac{1}{4} - D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.2.1.1.2.2

Factor $2$ out of $4$ .

$A = 2 \cdot (- 1 \frac{1}{2 \cdot 2}) - 2 (- D) - 2 D$

$C = \frac{1}{4} - D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.2.1.1.2.3

Cancel the common factor.

$A = 2 \cdot (- 1 \frac{1}{2 \cdot 2}) - 2 (- D) - 2 D$

$C = \frac{1}{4} - D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.2.1.1.2.4

Rewrite the expression.

$A = - 1 (\frac{1}{2}) - 2 (- D) - 2 D$

$C = \frac{1}{4} - D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$A = - 1 (\frac{1}{2}) - 2 (- D) - 2 D$

$C = \frac{1}{4} - D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.2.1.1.3

Multiply $- 1$ by $- 2$ .

$A = - 1 (\frac{1}{2}) + 2 D - 2 D$

$C = \frac{1}{4} - D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.2.1.1.4

Rewrite $- 1 (\frac{1}{2})$ as $- (\frac{1}{2})$ .

$A = - \frac{1}{2} + 2 D - 2 D$

$C = \frac{1}{4} - D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$A = - \frac{1}{2} + 2 D - 2 D$

$C = \frac{1}{4} - D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.2.1.2

Combine the opposite terms in $- \frac{1}{2} + 2 D - 2 D$ .

Tap for more steps...

Step 1.3.4.2.1.2.1

Subtract $2 D$ from $2 D$ .

$A = - \frac{1}{2} + 0$

$C = \frac{1}{4} - D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.2.1.2.2

Add $- \frac{1}{2}$ and $0$ .

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = 4 B - 4 C + 4 D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.3

Replace all occurrences of $C$ in $0 = 4 B - 4 C + 4 D$ with $\frac{1}{4} - D$ .

$0 = 4 B - 4 (\frac{1}{4} - D) + 4 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.4

Simplify the right side.

Tap for more steps...

Step 1.3.4.4.1

Simplify $4 B - 4 (\frac{1}{4} - D) + 4 D$ .

Tap for more steps...

Step 1.3.4.4.1.1

Simplify each term.

Tap for more steps...

Step 1.3.4.4.1.1.1

Apply the distributive property.

$0 = 4 B - 4 (\frac{1}{4}) - 4 (- D) + 4 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.4.1.1.2

Cancel the common factor of $4$ .

Tap for more steps...

Step 1.3.4.4.1.1.2.1

Factor $4$ out of $- 4$ .

$0 = 4 B + 4 (- 1) (\frac{1}{4}) - 4 (- D) + 4 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.4.1.1.2.2

Cancel the common factor.

$0 = 4 B + 4 \cdot (- 1 (\frac{1}{4})) - 4 (- D) + 4 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.4.1.1.2.3

Rewrite the expression.

$0 = 4 B - 1 - 4 (- D) + 4 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - 1 B - 1 C + D$

$0 = 4 B - 1 - 4 (- D) + 4 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.4.1.1.3

Multiply $- 1$ by $- 4$ .

$0 = 4 B - 1 + 4 D + 4 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - 1 B - 1 C + D$

$0 = 4 B - 1 + 4 D + 4 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.4.1.2

Add $4 D$ and $4 D$ .

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - 1 B - 1 C + D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - 1 B - 1 C + D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - 1 B - 1 C + D$

Step 1.3.4.5

Replace all occurrences of $C$ in $0 = - 1 B - 1 C + D$ with $\frac{1}{4} - D$ .

$0 = - 1 B - 1 (\frac{1}{4} - D) + D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

Step 1.3.4.6

Simplify the right side.

Tap for more steps...

Step 1.3.4.6.1

Simplify $- 1 B - 1 (\frac{1}{4} - D) + D$ .

Tap for more steps...

Step 1.3.4.6.1.1

Simplify each term.

Tap for more steps...

Step 1.3.4.6.1.1.1

Rewrite $- 1 B$ as $- B$ .

$0 = - B - 1 (\frac{1}{4} - D) + D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

Step 1.3.4.6.1.1.2

Apply the distributive property.

$0 = - B - 1 (\frac{1}{4}) - 1 (- D) + D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

Step 1.3.4.6.1.1.3

Rewrite $- 1 (\frac{1}{4})$ as $- (\frac{1}{4})$ .

$0 = - B - (\frac{1}{4}) - 1 (- D) + D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

Step 1.3.4.6.1.1.4

Multiply $- 1 (- D)$ .

Tap for more steps...

Step 1.3.4.6.1.1.4.1

Multiply $- 1$ by $- 1$ .

$0 = - B - (\frac{1}{4}) + 1 D + D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

Step 1.3.4.6.1.1.4.2

Multiply $D$ by $1$ .

$0 = - B - (\frac{1}{4}) + D + D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - B - \frac{1}{4} + D + D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - B - \frac{1}{4} + D + D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

Step 1.3.4.6.1.2

Add $D$ and $D$ .

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$C = \frac{1}{4} - D$

Step 1.3.5

Reorder $\frac{1}{4}$ and $- D$ .

$C = - D + \frac{1}{4}$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.6

Solve for $D$ in $C = - D + \frac{1}{4}$ .

Tap for more steps...

Step 1.3.6.1

Rewrite the equation as $- D + \frac{1}{4} = C$ .

$- D + \frac{1}{4} = C$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.6.2

Subtract $\frac{1}{4}$ from both sides of the equation.

$- D = C - \frac{1}{4}$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.6.3

Divide each term in $- D = C - \frac{1}{4}$ by $- 1$ and simplify.

Tap for more steps...

Step 1.3.6.3.1

Divide each term in $- D = C - \frac{1}{4}$ by $- 1$ .

$\frac{- D}{- 1} = \frac{C}{- 1} + \frac{- \frac{1}{4}}{- 1}$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.6.3.2

Simplify the left side.

Tap for more steps...

Step 1.3.6.3.2.1

Dividing two negative values results in a positive value.

$\frac{D}{1} = \frac{C}{- 1} + \frac{- \frac{1}{4}}{- 1}$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.6.3.2.2

Divide $D$ by $1$ .

$D = \frac{C}{- 1} + \frac{- \frac{1}{4}}{- 1}$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$D = \frac{C}{- 1} + \frac{- \frac{1}{4}}{- 1}$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.6.3.3

Simplify the right side.

Tap for more steps...

Step 1.3.6.3.3.1

Simplify each term.

Tap for more steps...

Step 1.3.6.3.3.1.1

Move the negative one from the denominator of $\frac{C}{- 1}$ .

$D = - 1 \cdot C + \frac{- \frac{1}{4}}{- 1}$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.6.3.3.1.2

Rewrite $- 1 \cdot C$ as $- C$ .

$D = - C + \frac{- \frac{1}{4}}{- 1}$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.6.3.3.1.3

Dividing two negative values results in a positive value.

$D = - C + \frac{\frac{1}{4}}{1}$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.6.3.3.1.4

Divide $\frac{1}{4}$ by $1$ .

$D = - C + \frac{1}{4}$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$D = - C + \frac{1}{4}$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$D = - C + \frac{1}{4}$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$D = - C + \frac{1}{4}$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$D = - C + \frac{1}{4}$

$0 = - B - \frac{1}{4} + 2 D$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.7

Replace all occurrences of $D$ with $- C + \frac{1}{4}$ in each equation.

Tap for more steps...

Step 1.3.7.1

Replace all occurrences of $D$ in $0 = - B - \frac{1}{4} + 2 D$ with $- C + \frac{1}{4}$ .

$0 = - B - \frac{1}{4} + 2 (- C + \frac{1}{4})$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.7.2

Simplify the right side.

Tap for more steps...

Step 1.3.7.2.1

Simplify $- B - \frac{1}{4} + 2 (- C + \frac{1}{4})$ .

Tap for more steps...

Step 1.3.7.2.1.1

Simplify each term.

Tap for more steps...

Step 1.3.7.2.1.1.1

Apply the distributive property.

$0 = - B - \frac{1}{4} + 2 (- C) + 2 (\frac{1}{4})$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.7.2.1.1.2

Multiply $- 1$ by $2$ .

$0 = - B - \frac{1}{4} - 2 C + 2 (\frac{1}{4})$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.7.2.1.1.3

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.3.7.2.1.1.3.1

Factor $2$ out of $4$ .

$0 = - B - \frac{1}{4} - 2 C + 2 (\frac{1}{2 (2)})$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.7.2.1.1.3.2

Cancel the common factor.

$0 = - B - \frac{1}{4} - 2 C + 2 (\frac{1}{2 \cdot 2})$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.7.2.1.1.3.3

Rewrite the expression.

$0 = - B - \frac{1}{4} - 2 C + \frac{1}{2}$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$0 = - B - \frac{1}{4} - 2 C + \frac{1}{2}$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$0 = - B - \frac{1}{4} - 2 C + \frac{1}{2}$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.7.2.1.2

To write $\frac{1}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$0 = - B - 2 C - \frac{1}{4} + \frac{1}{2} \cdot \frac{2}{2}$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.7.2.1.3

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

Tap for more steps...

Step 1.3.7.2.1.3.1

Multiply $\frac{1}{2}$ by $\frac{2}{2}$ .

$0 = - B - 2 C - \frac{1}{4} + \frac{2}{2 \cdot 2}$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.7.2.1.3.2

Multiply $2$ by $2$ .

$0 = - B - 2 C - \frac{1}{4} + \frac{2}{4}$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$0 = - B - 2 C - \frac{1}{4} + \frac{2}{4}$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.7.2.1.4

Combine the numerators over the common denominator.

$0 = - B - 2 C + \frac{- 1 + 2}{4}$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.7.2.1.5

Add $- 1$ and $2$ .

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$0 = 4 B - 1 + 8 D$

$A = - \frac{1}{2}$

Step 1.3.7.3

Replace all occurrences of $D$ in $0 = 4 B - 1 + 8 D$ with $- C + \frac{1}{4}$ .

$0 = 4 B - 1 + 8 (- C + \frac{1}{4})$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.7.4

Simplify the right side.

Tap for more steps...

Step 1.3.7.4.1

Simplify $4 B - 1 + 8 (- C + \frac{1}{4})$ .

Tap for more steps...

Step 1.3.7.4.1.1

Simplify each term.

Tap for more steps...

Step 1.3.7.4.1.1.1

Apply the distributive property.

$0 = 4 B - 1 + 8 (- C) + 8 (\frac{1}{4})$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.7.4.1.1.2

Multiply $- 1$ by $8$ .

$0 = 4 B - 1 - 8 C + 8 (\frac{1}{4})$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.7.4.1.1.3

Cancel the common factor of $4$ .

Tap for more steps...

Step 1.3.7.4.1.1.3.1

Factor $4$ out of $8$ .

$0 = 4 B - 1 - 8 C + 4 (2) (\frac{1}{4})$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.7.4.1.1.3.2

Cancel the common factor.

$0 = 4 B - 1 - 8 C + 4 \cdot (2 (\frac{1}{4}))$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.7.4.1.1.3.3

Rewrite the expression.

$0 = 4 B - 1 - 8 C + 2$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$0 = 4 B - 1 - 8 C + 2$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$0 = 4 B - 1 - 8 C + 2$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.7.4.1.2

Add $- 1$ and $2$ .

$0 = 4 B - 8 C + 1$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$0 = 4 B - 8 C + 1$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$0 = 4 B - 8 C + 1$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$0 = 4 B - 8 C + 1$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.8

Solve for $B$ in $0 = 4 B - 8 C + 1$ .

Tap for more steps...

Step 1.3.8.1

Rewrite the equation as $4 B - 8 C + 1 = 0$ .

$4 B - 8 C + 1 = 0$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.8.2

Move all terms not containing $B$ to the right side of the equation.

Tap for more steps...

Step 1.3.8.2.1

Add $8 C$ to both sides of the equation.

$4 B + 1 = 8 C$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.8.2.2

Subtract $1$ from both sides of the equation.

$4 B = 8 C - 1$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$4 B = 8 C - 1$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.8.3

Divide each term in $4 B = 8 C - 1$ by $4$ and simplify.

Tap for more steps...

Step 1.3.8.3.1

Divide each term in $4 B = 8 C - 1$ by $4$ .

$\frac{4 B}{4} = \frac{8 C}{4} + \frac{- 1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.8.3.2

Simplify the left side.

Tap for more steps...

Step 1.3.8.3.2.1

Cancel the common factor of $4$ .

Tap for more steps...

Step 1.3.8.3.2.1.1

Cancel the common factor.

$\frac{4 B}{4} = \frac{8 C}{4} + \frac{- 1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.8.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{8 C}{4} + \frac{- 1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$B = \frac{8 C}{4} + \frac{- 1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$B = \frac{8 C}{4} + \frac{- 1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.8.3.3

Simplify the right side.

Tap for more steps...

Step 1.3.8.3.3.1

Simplify each term.

Tap for more steps...

Step 1.3.8.3.3.1.1

Cancel the common factor of $8$ and $4$ .

Tap for more steps...

Step 1.3.8.3.3.1.1.1

Factor $4$ out of $8 C$ .

$B = \frac{4 (2 C)}{4} + \frac{- 1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.8.3.3.1.1.2

Cancel the common factors.

Tap for more steps...

Step 1.3.8.3.3.1.1.2.1

Factor $4$ out of $4$ .

$B = \frac{4 (2 C)}{4 (1)} + \frac{- 1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.8.3.3.1.1.2.2

Cancel the common factor.

$B = \frac{4 (2 C)}{4 \cdot 1} + \frac{- 1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.8.3.3.1.1.2.3

Rewrite the expression.

$B = \frac{2 C}{1} + \frac{- 1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.8.3.3.1.1.2.4

Divide $2 C$ by $1$ .

$B = 2 C + \frac{- 1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$B = 2 C + \frac{- 1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$B = 2 C + \frac{- 1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.8.3.3.1.2

Move the negative in front of the fraction.

$B = 2 C - \frac{1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$B = 2 C - \frac{1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$B = 2 C - \frac{1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$B = 2 C - \frac{1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$B = 2 C - \frac{1}{4}$

$0 = - B - 2 C + \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.9

Replace all occurrences of $B$ with $2 C - \frac{1}{4}$ in each equation.

Tap for more steps...

Step 1.3.9.1

Replace all occurrences of $B$ in $0 = - B - 2 C + \frac{1}{4}$ with $2 C - \frac{1}{4}$ .

$0 = - (2 C - \frac{1}{4}) - 2 C + \frac{1}{4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.9.2

Simplify the right side.

Tap for more steps...

Step 1.3.9.2.1

Simplify $- (2 C - \frac{1}{4}) - 2 C + \frac{1}{4}$ .

Tap for more steps...

Step 1.3.9.2.1.1

Simplify each term.

Tap for more steps...

Step 1.3.9.2.1.1.1

Apply the distributive property.

$0 = - (2 C) + \frac{1}{4} - 2 C + \frac{1}{4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.9.2.1.1.2

Multiply $2$ by $- 1$ .

$0 = - 2 C + \frac{1}{4} - 2 C + \frac{1}{4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.9.2.1.1.3

Multiply $- - \frac{1}{4}$ .

Tap for more steps...

Step 1.3.9.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$0 = - 2 C + 1 (\frac{1}{4}) - 2 C + \frac{1}{4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.9.2.1.1.3.2

Multiply $\frac{1}{4}$ by $1$ .

$0 = - 2 C + \frac{1}{4} - 2 C + \frac{1}{4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$0 = - 2 C + \frac{1}{4} - 2 C + \frac{1}{4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$0 = - 2 C + \frac{1}{4} - 2 C + \frac{1}{4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.9.2.1.2

Simplify terms.

Tap for more steps...

Step 1.3.9.2.1.2.1

Combine the numerators over the common denominator.

$0 = - 2 C - 2 C + \frac{1 + 1}{4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.9.2.1.2.2

Add $1$ and $1$ .

$0 = - 2 C - 2 C + \frac{2}{4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.9.2.1.2.3

Cancel the common factor of $2$ and $4$ .

Tap for more steps...

Step 1.3.9.2.1.2.3.1

Factor $2$ out of $2$ .

$0 = - 2 C - 2 C + \frac{2 (1)}{4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.9.2.1.2.3.2

Cancel the common factors.

Tap for more steps...

Step 1.3.9.2.1.2.3.2.1

Factor $2$ out of $4$ .

$0 = - 2 C - 2 C + \frac{2 \cdot 1}{2 \cdot 2}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.9.2.1.2.3.2.2

Cancel the common factor.

$0 = - 2 C - 2 C + \frac{2 \cdot 1}{2 \cdot 2}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.9.2.1.2.3.2.3

Rewrite the expression.

$0 = - 2 C - 2 C + \frac{1}{2}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$0 = - 2 C - 2 C + \frac{1}{2}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$0 = - 2 C - 2 C + \frac{1}{2}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.9.2.1.2.4

Subtract $2 C$ from $- 2 C$ .

$0 = - 4 C + \frac{1}{2}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$0 = - 4 C + \frac{1}{2}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$0 = - 4 C + \frac{1}{2}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$0 = - 4 C + \frac{1}{2}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$0 = - 4 C + \frac{1}{2}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.10

Solve for $C$ in $0 = - 4 C + \frac{1}{2}$ .

Tap for more steps...

Step 1.3.10.1

Rewrite the equation as $- 4 C + \frac{1}{2} = 0$ .

$- 4 C + \frac{1}{2} = 0$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.10.2

Subtract $\frac{1}{2}$ from both sides of the equation.

$- 4 C = - \frac{1}{2}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.10.3

Divide each term in $- 4 C = - \frac{1}{2}$ by $- 4$ and simplify.

Tap for more steps...

Step 1.3.10.3.1

Divide each term in $- 4 C = - \frac{1}{2}$ by $- 4$ .

$\frac{- 4 C}{- 4} = \frac{- \frac{1}{2}}{- 4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.10.3.2

Simplify the left side.

Tap for more steps...

Step 1.3.10.3.2.1

Cancel the common factor of $- 4$ .

Tap for more steps...

Step 1.3.10.3.2.1.1

Cancel the common factor.

$\frac{- 4 C}{- 4} = \frac{- \frac{1}{2}}{- 4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.10.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{- \frac{1}{2}}{- 4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$C = \frac{- \frac{1}{2}}{- 4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$C = \frac{- \frac{1}{2}}{- 4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.10.3.3

Simplify the right side.

Tap for more steps...

Step 1.3.10.3.3.1

Multiply the numerator by the reciprocal of the denominator.

$C = - \frac{1}{2} \cdot \frac{1}{- 4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.10.3.3.2

Move the negative in front of the fraction.

$C = - \frac{1}{2} \cdot (- \frac{1}{4})$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.10.3.3.3

Multiply $- \frac{1}{2} (- \frac{1}{4})$ .

Tap for more steps...

Step 1.3.10.3.3.3.1

Multiply $- 1$ by $- 1$ .

$C = 1 (\frac{1}{2}) \cdot \frac{1}{4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.10.3.3.3.2

Multiply $\frac{1}{2}$ by $1$ .

$C = \frac{1}{2} \cdot \frac{1}{4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.10.3.3.3.3

Multiply $\frac{1}{2}$ by $\frac{1}{4}$ .

$C = \frac{1}{2 \cdot 4}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.10.3.3.3.4

Multiply $2$ by $4$ .

$C = \frac{1}{8}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$C = \frac{1}{8}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$C = \frac{1}{8}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$C = \frac{1}{8}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$C = \frac{1}{8}$

$B = 2 C - \frac{1}{4}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.11

Replace all occurrences of $C$ with $\frac{1}{8}$ in each equation.

Tap for more steps...

Step 1.3.11.1

Replace all occurrences of $C$ in $B = 2 C - \frac{1}{4}$ with $\frac{1}{8}$ .

$B = 2 (\frac{1}{8}) - \frac{1}{4}$

$C = \frac{1}{8}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.11.2

Simplify the right side.

Tap for more steps...

Step 1.3.11.2.1

Simplify $2 (\frac{1}{8}) - \frac{1}{4}$ .

Tap for more steps...

Step 1.3.11.2.1.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.3.11.2.1.1.1

Factor $2$ out of $8$ .

$B = 2 (\frac{1}{2 (4)}) - \frac{1}{4}$

$C = \frac{1}{8}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.11.2.1.1.2

Cancel the common factor.

$B = 2 (\frac{1}{2 \cdot 4}) - \frac{1}{4}$

$C = \frac{1}{8}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.11.2.1.1.3

Rewrite the expression.

$B = \frac{1}{4} - \frac{1}{4}$

$C = \frac{1}{8}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$B = \frac{1}{4} - \frac{1}{4}$

$C = \frac{1}{8}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.11.2.1.2

Combine the numerators over the common denominator.

$B = \frac{1 - 1}{4}$

$C = \frac{1}{8}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.11.2.1.3

Simplify the expression.

Tap for more steps...

Step 1.3.11.2.1.3.1

Subtract $1$ from $1$ .

$B = \frac{0}{4}$

$C = \frac{1}{8}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.11.2.1.3.2

Divide $0$ by $4$ .

$B = 0$

$C = \frac{1}{8}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$B = 0$

$C = \frac{1}{8}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$B = 0$

$C = \frac{1}{8}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

$B = 0$

$C = \frac{1}{8}$

$D = - C + \frac{1}{4}$

$A = - \frac{1}{2}$

Step 1.3.11.3

Replace all occurrences of $C$ in $D = - C + \frac{1}{4}$ with $\frac{1}{8}$ .

$D = - (\frac{1}{8}) + \frac{1}{4}$

$B = 0$

$C = \frac{1}{8}$

$A = - \frac{1}{2}$

Step 1.3.11.4

Simplify the right side.

Tap for more steps...

Step 1.3.11.4.1

Simplify $- (\frac{1}{8}) + \frac{1}{4}$ .

Tap for more steps...

Step 1.3.11.4.1.1

To write $\frac{1}{4}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$D = - \frac{1}{8} + \frac{1}{4} \cdot \frac{2}{2}$

$B = 0$

$C = \frac{1}{8}$

$A = - \frac{1}{2}$

Step 1.3.11.4.1.2

Write each expression with a common denominator of $8$ , by multiplying each by an appropriate factor of $1$ .

Tap for more steps...

Step 1.3.11.4.1.2.1

Multiply $\frac{1}{4}$ by $\frac{2}{2}$ .

$D = - \frac{1}{8} + \frac{2}{4 \cdot 2}$

$B = 0$

$C = \frac{1}{8}$

$A = - \frac{1}{2}$

Step 1.3.11.4.1.2.2

Multiply $4$ by $2$ .

$D = - \frac{1}{8} + \frac{2}{8}$

$B = 0$

$C = \frac{1}{8}$

$A = - \frac{1}{2}$

$D = - \frac{1}{8} + \frac{2}{8}$

$B = 0$

$C = \frac{1}{8}$

$A = - \frac{1}{2}$

Step 1.3.11.4.1.3

Combine the numerators over the common denominator.

$D = \frac{- 1 + 2}{8}$

$B = 0$

$C = \frac{1}{8}$

$A = - \frac{1}{2}$

Step 1.3.11.4.1.4

Add $- 1$ and $2$ .

$D = \frac{1}{8}$

$B = 0$

$C = \frac{1}{8}$

$A = - \frac{1}{2}$

$D = \frac{1}{8}$

$B = 0$

$C = \frac{1}{8}$

$A = - \frac{1}{2}$

$D = \frac{1}{8}$

$B = 0$

$C = \frac{1}{8}$

$A = - \frac{1}{2}$

$D = \frac{1}{8}$

$B = 0$

$C = \frac{1}{8}$

$A = - \frac{1}{2}$

Step 1.3.12

List all of the solutions.

$D = \frac{1}{8}, B = 0, C = \frac{1}{8}, A = - \frac{1}{2}$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A x + B}{4 x^{2} + 1} + \frac{C}{2 x + 1} + \frac{D}{2 x - 1}$ with the values found for $A$ , $B$ , $C$ , and $D$ .

$\frac{- \frac{1}{2 x} + 0}{4 x^{2} + 1} + \frac{\frac{1}{8}}{2 x + 1} + \frac{\frac{1}{8}}{2 x - 1}$

Step 1.5

Simplify.

Tap for more steps...

Step 1.5.1

Simplify the numerator.

Tap for more steps...

Step 1.5.1.1

Combine $x$ and $\frac{1}{2}$ .

$\frac{- \frac{x}{2} + 0}{4 x^{2} + 1} + \frac{\frac{1}{8}}{2 x + 1} + \frac{\frac{1}{8}}{2 x - 1}$

Step 1.5.1.2

Add $- \frac{x}{2}$ and $0$ .

$\frac{- \frac{x}{2}}{4 x^{2} + 1} + \frac{\frac{1}{8}}{2 x + 1} + \frac{\frac{1}{8}}{2 x - 1}$

Step 1.5.2

Multiply the numerator by the reciprocal of the denominator.

$- \frac{x}{2} \cdot \frac{1}{4 x^{2} + 1} + \frac{\frac{1}{8}}{2 x + 1} + \frac{\frac{1}{8}}{2 x - 1}$

Step 1.5.3

Multiply $\frac{1}{4 x^{2} + 1}$ by $\frac{x}{2}$ .

$- \frac{x}{(4 x^{2} + 1) \cdot 2} + \frac{\frac{1}{8}}{2 x + 1} + \frac{\frac{1}{8}}{2 x - 1}$

Step 1.5.4

Move $2$ to the left of $4 x^{2} + 1$ .

$- \frac{x}{2 (4 x^{2} + 1)} + \frac{\frac{1}{8}}{2 x + 1} + \frac{\frac{1}{8}}{2 x - 1}$

Step 1.5.5

Multiply the numerator by the reciprocal of the denominator.

$- \frac{x}{2 (4 x^{2} + 1)} + \frac{1}{8} \cdot \frac{1}{2 x + 1} + \frac{\frac{1}{8}}{2 x - 1}$

Step 1.5.6

Multiply $\frac{1}{8}$ by $\frac{1}{2 x + 1}$ .

$- \frac{x}{2 (4 x^{2} + 1)} + \frac{1}{8 (2 x + 1)} + \frac{\frac{1}{8}}{2 x - 1}$

Step 1.5.7

Multiply the numerator by the reciprocal of the denominator.

$- \frac{x}{2 (4 x^{2} + 1)} + \frac{1}{8 (2 x + 1)} + \frac{1}{8} \cdot \frac{1}{2 x - 1}$

Step 1.5.8

Multiply $\frac{1}{8}$ by $\frac{1}{2 x - 1}$ .

$\int - \frac{x}{2 (4 x^{2} + 1)} + \frac{1}{8 (2 x + 1)} + \frac{1}{8 (2 x - 1)} d x$

Step 2

Split the single integral into multiple integrals.

$\int - \frac{x}{2 (4 x^{2} + 1)} d x + \int \frac{1}{8 (2 x + 1)} d x + \int \frac{1}{8 (2 x - 1)} d x$

Step 3

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$- \int \frac{x}{2 (4 x^{2} + 1)} d x + \int \frac{1}{8 (2 x + 1)} d x + \int \frac{1}{8 (2 x - 1)} d x$

Step 4

Since $\frac{1}{2}$ is constant with respect to $x$ , move $\frac{1}{2}$ out of the integral.

$- (\frac{1}{2} \int \frac{x}{4 x^{2} + 1} d x) + \int \frac{1}{8 (2 x + 1)} d x + \int \frac{1}{8 (2 x - 1)} d x$

Step 5

Let $u_{1} = 4 x^{2} + 1$ . Then $d u_{1} = 8 x d x$ , so $\frac{1}{8} d u_{1} = x d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

Tap for more steps...

Step 5.1

Let $u_{1} = 4 x^{2} + 1$ . Find $\frac{d u_{1}}{d x}$ .

Tap for more steps...

Step 5.1.1

Differentiate $4 x^{2} + 1$ .

$\frac{d}{d x} [4 x^{2} + 1]$

Step 5.1.2

By the Sum Rule, the derivative of $4 x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [1]$

Step 5.1.3

Evaluate $\frac{d}{d x} [4 x^{2}]$ .

Tap for more steps...

Step 5.1.3.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2}]$ .

$4 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$

Step 5.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$4 (2 x) + \frac{d}{d x} [1]$

Step 5.1.3.3

Multiply $2$ by $4$ .

$8 x + \frac{d}{d x} [1]$

Step 5.1.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 5.1.4.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$8 x + 0$

Step 5.1.4.2

Add $8 x$ and $0$ .

$8 x$

Step 5.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$- \frac{1}{2} \int \frac{1}{u_{1}} \cdot \frac{1}{8} d u_{1} + \int \frac{1}{8 (2 x + 1)} d x + \int \frac{1}{8 (2 x - 1)} d x$

Step 6

Simplify.

Tap for more steps...

Step 6.1

Multiply $\frac{1}{u_{1}}$ by $\frac{1}{8}$ .

$- \frac{1}{2} \int \frac{1}{u_{1} \cdot 8} d u_{1} + \int \frac{1}{8 (2 x + 1)} d x + \int \frac{1}{8 (2 x - 1)} d x$

Step 6.2

Move $8$ to the left of $u_{1}$ .

$- \frac{1}{2} \int \frac{1}{8 u_{1}} d u_{1} + \int \frac{1}{8 (2 x + 1)} d x + \int \frac{1}{8 (2 x - 1)} d x$

Step 7

Since $\frac{1}{8}$ is constant with respect to $u_{1}$ , move $\frac{1}{8}$ out of the integral.

$- \frac{1}{2} (\frac{1}{8} \int \frac{1}{u_{1}} d u_{1}) + \int \frac{1}{8 (2 x + 1)} d x + \int \frac{1}{8 (2 x - 1)} d x$

Step 8

Simplify.

Tap for more steps...

Step 8.1

Multiply $\frac{1}{8}$ by $\frac{1}{2}$ .

$- \frac{1}{8 \cdot 2} \int \frac{1}{u_{1}} d u_{1} + \int \frac{1}{8 (2 x + 1)} d x + \int \frac{1}{8 (2 x - 1)} d x$

Step 8.2

Multiply $8$ by $2$ .

$- \frac{1}{16} \int \frac{1}{u_{1}} d u_{1} + \int \frac{1}{8 (2 x + 1)} d x + \int \frac{1}{8 (2 x - 1)} d x$

Step 9

The integral of $\frac{1}{u_{1}}$ with respect to $u_{1}$ is $\ln (| u_{1} |)$ .

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \int \frac{1}{8 (2 x + 1)} d x + \int \frac{1}{8 (2 x - 1)} d x$

Step 10

Since $\frac{1}{8}$ is constant with respect to $x$ , move $\frac{1}{8}$ out of the integral.

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{8} \int \frac{1}{2 x + 1} d x + \int \frac{1}{8 (2 x - 1)} d x$

Step 11

Let $u_{2} = 2 x + 1$ . Then $d u_{2} = 2 d x$ , so $\frac{1}{2} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

Tap for more steps...

Step 11.1

Let $u_{2} = 2 x + 1$ . Find $\frac{d u_{2}}{d x}$ .

Tap for more steps...

Step 11.1.1

Differentiate $2 x + 1$ .

$\frac{d}{d x} [2 x + 1]$

Step 11.1.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$

Step 11.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

Tap for more steps...

Step 11.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 11.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$2 \cdot 1 + \frac{d}{d x} [1]$

Step 11.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [1]$

Step 11.1.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 11.1.4.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$2 + 0$

Step 11.1.4.2

Add $2$ and $0$ .

$2$

Step 11.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{8} \int \frac{1}{u_{2}} \cdot \frac{1}{2} d u_{2} + \int \frac{1}{8 (2 x - 1)} d x$

Step 12

Simplify.

Tap for more steps...

Step 12.1

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{2}$ .

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{8} \int \frac{1}{u_{2} \cdot 2} d u_{2} + \int \frac{1}{8 (2 x - 1)} d x$

Step 12.2

Move $2$ to the left of $u_{2}$ .

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{8} \int \frac{1}{2 u_{2}} d u_{2} + \int \frac{1}{8 (2 x - 1)} d x$

Step 13

Since $\frac{1}{2}$ is constant with respect to $u_{2}$ , move $\frac{1}{2}$ out of the integral.

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{8} (\frac{1}{2} \int \frac{1}{u_{2}} d u_{2}) + \int \frac{1}{8 (2 x - 1)} d x$

Step 14

Simplify.

Tap for more steps...

Step 14.1

Multiply $\frac{1}{2}$ by $\frac{1}{8}$ .

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{2 \cdot 8} \int \frac{1}{u_{2}} d u_{2} + \int \frac{1}{8 (2 x - 1)} d x$

Step 14.2

Multiply $2$ by $8$ .

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{16} \int \frac{1}{u_{2}} d u_{2} + \int \frac{1}{8 (2 x - 1)} d x$

Step 15

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{16} (\ln (| u_{2} |) + C) + \int \frac{1}{8 (2 x - 1)} d x$

Step 16

Since $\frac{1}{8}$ is constant with respect to $x$ , move $\frac{1}{8}$ out of the integral.

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{16} (\ln (| u_{2} |) + C) + \frac{1}{8} \int \frac{1}{2 x - 1} d x$

Step 17

Let $u_{3} = 2 x - 1$ . Then $d u_{3} = 2 d x$ , so $\frac{1}{2} d u_{3} = d x$ . Rewrite using $u_{3}$ and $d$ $u_{3}$ .

Tap for more steps...

Step 17.1

Let $u_{3} = 2 x - 1$ . Find $\frac{d u_{3}}{d x}$ .

Tap for more steps...

Step 17.1.1

Differentiate $2 x - 1$ .

$\frac{d}{d x} [2 x - 1]$

Step 17.1.2

By the Sum Rule, the derivative of $2 x - 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$

Step 17.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

Tap for more steps...

Step 17.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

Step 17.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$2 \cdot 1 + \frac{d}{d x} [- 1]$

Step 17.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [- 1]$

Step 17.1.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 17.1.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$2 + 0$

Step 17.1.4.2

Add $2$ and $0$ .

$2$

Step 17.2

Rewrite the problem using $u_{3}$ and $d u_{3}$ .

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{16} (\ln (| u_{2} |) + C) + \frac{1}{8} \int \frac{1}{u_{3}} \cdot \frac{1}{2} d u_{3}$

Step 18

Simplify.

Tap for more steps...

Step 18.1

Multiply $\frac{1}{u_{3}}$ by $\frac{1}{2}$ .

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{16} (\ln (| u_{2} |) + C) + \frac{1}{8} \int \frac{1}{u_{3} \cdot 2} d u_{3}$

Step 18.2

Move $2$ to the left of $u_{3}$ .

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{16} (\ln (| u_{2} |) + C) + \frac{1}{8} \int \frac{1}{2 u_{3}} d u_{3}$

Step 19

Since $\frac{1}{2}$ is constant with respect to $u_{3}$ , move $\frac{1}{2}$ out of the integral.

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{16} (\ln (| u_{2} |) + C) + \frac{1}{8} (\frac{1}{2} \int \frac{1}{u_{3}} d u_{3})$

Step 20

Simplify.

Tap for more steps...

Step 20.1

Multiply $\frac{1}{2}$ by $\frac{1}{8}$ .

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{16} (\ln (| u_{2} |) + C) + \frac{1}{2 \cdot 8} \int \frac{1}{u_{3}} d u_{3}$

Step 20.2

Multiply $2$ by $8$ .

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{16} (\ln (| u_{2} |) + C) + \frac{1}{16} \int \frac{1}{u_{3}} d u_{3}$

Step 21

The integral of $\frac{1}{u_{3}}$ with respect to $u_{3}$ is $\ln (| u_{3} |)$ .

$- \frac{1}{16} (\ln (| u_{1} |) + C) + \frac{1}{16} (\ln (| u_{2} |) + C) + \frac{1}{16} (\ln (| u_{3} |) + C)$

Step 22

Simplify.

$- \frac{1}{16} \ln (| u_{1} |) + \frac{1}{16} \ln (| u_{2} |) + \frac{1}{16} \ln (| u_{3} |) + C$

Step 23

Substitute back in for each integration substitution variable.

Tap for more steps...

Step 23.1

Replace all occurrences of $u_{1}$ with $4 x^{2} + 1$ .

$- \frac{1}{16} \ln (| 4 x^{2} + 1 |) + \frac{1}{16} \ln (| u_{2} |) + \frac{1}{16} \ln (| u_{3} |) + C$

Step 23.2

Replace all occurrences of $u_{2}$ with $2 x + 1$ .

$- \frac{1}{16} \ln (| 4 x^{2} + 1 |) + \frac{1}{16} \ln (| 2 x + 1 |) + \frac{1}{16} \ln (| u_{3} |) + C$

Step 23.3

Replace all occurrences of $u_{3}$ with $2 x - 1$ .

$- \frac{1}{16} \ln (| 4 x^{2} + 1 |) + \frac{1}{16} \ln (| 2 x + 1 |) + \frac{1}{16} \ln (| 2 x - 1 |) + C$