Calculus Examples

$\int \frac{x^{2} + 2 x - 1}{2 x^{3} + 3 x^{2} - 2 x} d x$

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

Factor $x$ out of $2 x^{3} + 3 x^{2} - 2 x$ .

Tap for more steps...

Step 1.1.1.1.1

Factor $x$ out of $2 x^{3}$ .

$\frac{x^{2} + 2 x - 1}{x (2 x^{2}) + 3 x^{2} - 2 x}$

Step 1.1.1.1.2

Factor $x$ out of $3 x^{2}$ .

$\frac{x^{2} + 2 x - 1}{x (2 x^{2}) + x (3 x) - 2 x}$

Step 1.1.1.1.3

Factor $x$ out of $- 2 x$ .

$\frac{x^{2} + 2 x - 1}{x (2 x^{2}) + x (3 x) + x \cdot - 2}$

Step 1.1.1.1.4

Factor $x$ out of $x (2 x^{2}) + x (3 x)$ .

$\frac{x^{2} + 2 x - 1}{x (2 x^{2} + 3 x) + x \cdot - 2}$

Step 1.1.1.1.5

Factor $x$ out of $x (2 x^{2} + 3 x) + x \cdot - 2$ .

$\frac{x^{2} + 2 x - 1}{x (2 x^{2} + 3 x - 2)}$

Step 1.1.1.2

Factor.

Tap for more steps...

Step 1.1.1.2.1

Factor by grouping.

Tap for more steps...

Step 1.1.1.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 2 = - 4$ and whose sum is $b = 3$ .

Tap for more steps...

Step 1.1.1.2.1.1.1

Factor $3$ out of $3 x$ .

$\frac{x^{2} + 2 x - 1}{x (2 x^{2} + 3 (x) - 2)}$

Step 1.1.1.2.1.1.2

Rewrite $3$ as $- 1$ plus $4$

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

Step 1.1.1.2.1.1.3

Apply the distributive property.

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

Step 1.1.1.2.1.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 1.1.1.2.1.2.1

Group the first two terms and the last two terms.

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

Step 1.1.1.2.1.2.2

Factor out the greatest common factor (GCF) from each group.

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

Step 1.1.1.2.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

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

Step 1.1.1.2.2

Remove unnecessary parentheses.

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

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 in the denominator is linear, put a single variable in its place $B$ .

$\frac{A}{x} + \frac{B}{2 x - 1}$

Step 1.1.3

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

Step 1.1.4

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

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

Step 1.1.5

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.1.5.1

Cancel the common factor.

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

Step 1.1.5.2

Rewrite the expression.

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

Step 1.1.6

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

Tap for more steps...

Step 1.1.6.1

Cancel the common factor.

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

Step 1.1.6.2

Rewrite the expression.

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

Step 1.1.7

Cancel the common factor of $x + 2$ .

Tap for more steps...

Step 1.1.7.1

Cancel the common factor.

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

Step 1.1.7.2

Divide $x^{2} + 2 x - 1$ by $1$ .

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

Step 1.1.8

Simplify each term.

Tap for more steps...

Step 1.1.8.1

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.1.8.1.1

Cancel the common factor.

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

Step 1.1.8.1.2

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

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

Step 1.1.8.2

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

Tap for more steps...

Step 1.1.8.2.1

Apply the distributive property.

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

Step 1.1.8.2.2

Apply the distributive property.

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

Step 1.1.8.2.3

Apply the distributive property.

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

Step 1.1.8.3

Simplify and combine like terms.

Tap for more steps...

Step 1.1.8.3.1

Simplify each term.

Tap for more steps...

Step 1.1.8.3.1.1

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

Tap for more steps...

Step 1.1.8.3.1.1.1

Move $x$ .

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

Step 1.1.8.3.1.1.2

Multiply $x$ by $x$ .

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

Step 1.1.8.3.1.2

Multiply $2$ by $2$ .

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

Step 1.1.8.3.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 1.1.8.3.1.4

Multiply $- 1$ by $2$ .

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

Step 1.1.8.3.2

Subtract $x$ from $4 x$ .

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

Step 1.1.8.4

Apply the distributive property.

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

Step 1.1.8.5

Simplify.

Tap for more steps...

Step 1.1.8.5.1

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.5.2

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.5.3

Move $- 2$ to the left of $A$ .

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

Step 1.1.8.6

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

Tap for more steps...

Step 1.1.8.6.1

Cancel the common factor.

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

Step 1.1.8.6.2

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

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

Step 1.1.8.7

Apply the distributive property.

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

Step 1.1.8.8

Multiply $x$ by $x$ .

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

Step 1.1.8.9

Move $2$ to the left of $x$ .

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

Step 1.1.8.10

Apply the distributive property.

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

Step 1.1.8.11

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.12

Cancel the common factor of $x + 2$ .

Tap for more steps...

Step 1.1.8.12.1

Cancel the common factor.

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

Step 1.1.8.12.2

Divide $(C) (x (2 x - 1))$ by $1$ .

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

Step 1.1.8.13

Apply the distributive property.

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

Step 1.1.8.14

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.15

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

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

Step 1.1.8.16

Simplify each term.

Tap for more steps...

Step 1.1.8.16.1

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

Tap for more steps...

Step 1.1.8.16.1.1

Move $x$ .

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

Step 1.1.8.16.1.2

Multiply $x$ by $x$ .

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

Step 1.1.8.16.2

Rewrite $- 1 x$ as $- x$ .

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

Step 1.1.8.17

Apply the distributive property.

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

Step 1.1.8.18

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.19

Rewrite using the commutative property of multiplication.

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

Step 1.1.9

Simplify the expression.

Tap for more steps...

Step 1.1.9.1

Move $A$ .

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

Step 1.1.9.2

Move $A$ .

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

Step 1.1.9.3

Reorder $B$ and $x^{2}$ .

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

Step 1.1.9.4

Move $B$ .

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

Step 1.1.9.5

Move $- 2 A$ .

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

Step 1.1.9.6

Move $2 x B$ .

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

Step 1.1.9.7

Move $3 x A$ .

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

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 $x^{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = 2 A + B + 2 C$

Step 1.2.2

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.

$2 = 3 A + 2 B - 1 C$

Step 1.2.3

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.

$- 1 = - 2 A$

Step 1.2.4

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

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

$- 1 = - 2 A$

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

$- 1 = - 2 A$

Step 1.3

Solve the system of equations.

Tap for more steps...

Step 1.3.1

Solve for $A$ in $- 1 = - 2 A$ .

Tap for more steps...

Step 1.3.1.1

Rewrite the equation as $- 2 A = - 1$ .

$- 2 A = - 1$

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

Step 1.3.1.2

Divide each term in $- 2 A = - 1$ by $- 2$ and simplify.

Tap for more steps...

Step 1.3.1.2.1

Divide each term in $- 2 A = - 1$ by $- 2$ .

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

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

Step 1.3.1.2.2

Simplify the left side.

Tap for more steps...

Step 1.3.1.2.2.1

Cancel the common factor of $- 2$ .

Tap for more steps...

Step 1.3.1.2.2.1.1

Cancel the common factor.

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

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

Step 1.3.1.2.2.1.2

Divide $A$ by $1$ .

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

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

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

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

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

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

Step 1.3.1.2.3

Simplify the right side.

Tap for more steps...

Step 1.3.1.2.3.1

Dividing two negative values results in a positive value.

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

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

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

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

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

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

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

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

Step 1.3.2

Replace all occurrences of $A$ with $\frac{1}{2}$ in each equation.

Tap for more steps...

Step 1.3.2.1

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

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

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

$2 = 3 A + 2 B - 1 C$

Step 1.3.2.2

Simplify the right side.

Tap for more steps...

Step 1.3.2.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.3.2.2.1.1

Cancel the common factor.

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

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

$2 = 3 A + 2 B - 1 C$

Step 1.3.2.2.1.2

Rewrite the expression.

$1 = 1 + B + 2 C$

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

$2 = 3 A + 2 B - 1 C$

$1 = 1 + B + 2 C$

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

$2 = 3 A + 2 B - 1 C$

$1 = 1 + B + 2 C$

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

$2 = 3 A + 2 B - 1 C$

Step 1.3.2.3

Replace all occurrences of $A$ in $2 = 3 A + 2 B - 1 C$ with $\frac{1}{2}$ .

$2 = 3 (\frac{1}{2}) + 2 B - 1 C$

$1 = 1 + B + 2 C$

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

Step 1.3.2.4

Simplify the right side.

Tap for more steps...

Step 1.3.2.4.1

Simplify each term.

Tap for more steps...

Step 1.3.2.4.1.1

Combine $3$ and $\frac{1}{2}$ .

$2 = \frac{3}{2} + 2 B - 1 C$

$1 = 1 + B + 2 C$

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

Step 1.3.2.4.1.2

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

$2 = \frac{3}{2} + 2 B - C$

$1 = 1 + B + 2 C$

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

$2 = \frac{3}{2} + 2 B - C$

$1 = 1 + B + 2 C$

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

$2 = \frac{3}{2} + 2 B - C$

$1 = 1 + B + 2 C$

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

$2 = \frac{3}{2} + 2 B - C$

$1 = 1 + B + 2 C$

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

Step 1.3.3

Solve for $B$ in $1 = 1 + B + 2 C$ .

Tap for more steps...

Step 1.3.3.1

Rewrite the equation as $1 + B + 2 C = 1$ .

$1 + B + 2 C = 1$

$2 = \frac{3}{2} + 2 B - C$

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

Step 1.3.3.2

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

Tap for more steps...

Step 1.3.3.2.1

Subtract $1$ from both sides of the equation.

$B + 2 C = 1 - 1$

$2 = \frac{3}{2} + 2 B - C$

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

Step 1.3.3.2.2

Subtract $2 C$ from both sides of the equation.

$B = 1 - 1 - 2 C$

$2 = \frac{3}{2} + 2 B - C$

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

Step 1.3.3.2.3

Subtract $1$ from $1$ .

$B = 0 - 2 C$

$2 = \frac{3}{2} + 2 B - C$

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

Step 1.3.3.2.4

Subtract $2 C$ from $0$ .

$B = - 2 C$

$2 = \frac{3}{2} + 2 B - C$

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

$B = - 2 C$

$2 = \frac{3}{2} + 2 B - C$

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

$B = - 2 C$

$2 = \frac{3}{2} + 2 B - C$

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

Step 1.3.4

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

Tap for more steps...

Step 1.3.4.1

Replace all occurrences of $B$ in $2 = \frac{3}{2} + 2 B - C$ with $- 2 C$ .

$2 = \frac{3}{2} + 2 (- 2 C) - C$

$B = - 2 C$

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

Step 1.3.4.2

Simplify the right side.

Tap for more steps...

Step 1.3.4.2.1

Simplify $\frac{3}{2} + 2 (- 2 C) - C$ .

Tap for more steps...

Step 1.3.4.2.1.1

Multiply $- 2$ by $2$ .

$2 = \frac{3}{2} - 4 C - C$

$B = - 2 C$

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

Step 1.3.4.2.1.2

Subtract $C$ from $- 4 C$ .

$2 = \frac{3}{2} - 5 C$

$B = - 2 C$

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

$2 = \frac{3}{2} - 5 C$

$B = - 2 C$

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

$2 = \frac{3}{2} - 5 C$

$B = - 2 C$

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

$2 = \frac{3}{2} - 5 C$

$B = - 2 C$

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

Step 1.3.5

Solve for $C$ in $2 = \frac{3}{2} - 5 C$ .

Tap for more steps...

Step 1.3.5.1

Rewrite the equation as $\frac{3}{2} - 5 C = 2$ .

$\frac{3}{2} - 5 C = 2$

$B = - 2 C$

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

Step 1.3.5.2

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

Tap for more steps...

Step 1.3.5.2.1

Subtract $\frac{3}{2}$ from both sides of the equation.

$- 5 C = 2 - \frac{3}{2}$

$B = - 2 C$

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

Step 1.3.5.2.2

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

$- 5 C = 2 \cdot \frac{2}{2} - \frac{3}{2}$

$B = - 2 C$

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

Step 1.3.5.2.3

Combine $2$ and $\frac{2}{2}$ .

$- 5 C = \frac{2 \cdot 2}{2} - \frac{3}{2}$

$B = - 2 C$

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

Step 1.3.5.2.4

Combine the numerators over the common denominator.

$- 5 C = \frac{2 \cdot 2 - 3}{2}$

$B = - 2 C$

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

Step 1.3.5.2.5

Simplify the numerator.

Tap for more steps...

Step 1.3.5.2.5.1

Multiply $2$ by $2$ .

$- 5 C = \frac{4 - 3}{2}$

$B = - 2 C$

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

Step 1.3.5.2.5.2

Subtract $3$ from $4$ .

$- 5 C = \frac{1}{2}$

$B = - 2 C$

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

$- 5 C = \frac{1}{2}$

$B = - 2 C$

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

$- 5 C = \frac{1}{2}$

$B = - 2 C$

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

Step 1.3.5.3

Divide each term in $- 5 C = \frac{1}{2}$ by $- 5$ and simplify.

Tap for more steps...

Step 1.3.5.3.1

Divide each term in $- 5 C = \frac{1}{2}$ by $- 5$ .

$\frac{- 5 C}{- 5} = \frac{\frac{1}{2}}{- 5}$

$B = - 2 C$

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

Step 1.3.5.3.2

Simplify the left side.

Tap for more steps...

Step 1.3.5.3.2.1

Cancel the common factor of $- 5$ .

Tap for more steps...

Step 1.3.5.3.2.1.1

Cancel the common factor.

$\frac{- 5 C}{- 5} = \frac{\frac{1}{2}}{- 5}$

$B = - 2 C$

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

Step 1.3.5.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{\frac{1}{2}}{- 5}$

$B = - 2 C$

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

$C = \frac{\frac{1}{2}}{- 5}$

$B = - 2 C$

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

$C = \frac{\frac{1}{2}}{- 5}$

$B = - 2 C$

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

Step 1.3.5.3.3

Simplify the right side.

Tap for more steps...

Step 1.3.5.3.3.1

Multiply the numerator by the reciprocal of the denominator.

$C = \frac{1}{2} \cdot \frac{1}{- 5}$

$B = - 2 C$

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

Step 1.3.5.3.3.2

Move the negative in front of the fraction.

$C = \frac{1}{2} \cdot (- \frac{1}{5})$

$B = - 2 C$

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

Step 1.3.5.3.3.3

Multiply $\frac{1}{2} (- \frac{1}{5})$ .

Tap for more steps...

Step 1.3.5.3.3.3.1

Multiply $\frac{1}{2}$ by $\frac{1}{5}$ .

$C = - \frac{1}{2 \cdot 5}$

$B = - 2 C$

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

Step 1.3.5.3.3.3.2

Multiply $2$ by $5$ .

$C = - \frac{1}{10}$

$B = - 2 C$

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

$C = - \frac{1}{10}$

$B = - 2 C$

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

$C = - \frac{1}{10}$

$B = - 2 C$

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

$C = - \frac{1}{10}$

$B = - 2 C$

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

$C = - \frac{1}{10}$

$B = - 2 C$

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

Step 1.3.6

Replace all occurrences of $C$ with $- \frac{1}{10}$ in each equation.

Tap for more steps...

Step 1.3.6.1

Replace all occurrences of $C$ in $B = - 2 C$ with $- \frac{1}{10}$ .

$B = - 2 (- \frac{1}{10})$

$C = - \frac{1}{10}$

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

Step 1.3.6.2

Simplify the right side.

Tap for more steps...

Step 1.3.6.2.1

Simplify $- 2 (- \frac{1}{10})$ .

Tap for more steps...

Step 1.3.6.2.1.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.3.6.2.1.1.1

Move the leading negative in $- \frac{1}{10}$ into the numerator.

$B = - 2 (\frac{- 1}{10})$

$C = - \frac{1}{10}$

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

Step 1.3.6.2.1.1.2

Factor $2$ out of $- 2$ .

$B = 2 (- 1) (\frac{- 1}{10})$

$C = - \frac{1}{10}$

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

Step 1.3.6.2.1.1.3

Factor $2$ out of $10$ .

$B = 2 \cdot (- 1 \frac{- 1}{2 \cdot 5})$

$C = - \frac{1}{10}$

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

Step 1.3.6.2.1.1.4

Cancel the common factor.

$B = 2 \cdot (- 1 \frac{- 1}{2 \cdot 5})$

$C = - \frac{1}{10}$

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

Step 1.3.6.2.1.1.5

Rewrite the expression.

$B = - 1 (\frac{- 1}{5})$

$C = - \frac{1}{10}$

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

$B = - 1 (\frac{- 1}{5})$

$C = - \frac{1}{10}$

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

Step 1.3.6.2.1.2

Move the negative in front of the fraction.

$B = - 1 (- \frac{1}{5})$

$C = - \frac{1}{10}$

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

Step 1.3.6.2.1.3

Multiply $- 1 (- \frac{1}{5})$ .

Tap for more steps...

Step 1.3.6.2.1.3.1

Multiply $- 1$ by $- 1$ .

$B = 1 (\frac{1}{5})$

$C = - \frac{1}{10}$

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

Step 1.3.6.2.1.3.2

Multiply $\frac{1}{5}$ by $1$ .

$B = \frac{1}{5}$

$C = - \frac{1}{10}$

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

$B = \frac{1}{5}$

$C = - \frac{1}{10}$

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

$B = \frac{1}{5}$

$C = - \frac{1}{10}$

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

$B = \frac{1}{5}$

$C = - \frac{1}{10}$

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

$B = \frac{1}{5}$

$C = - \frac{1}{10}$

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

Step 1.3.7

List all of the solutions.

$B = \frac{1}{5}, C = - \frac{1}{10}, A = \frac{1}{2}$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{2 x - 1} + \frac{C}{x + 2}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{\frac{1}{2}}{x} + \frac{\frac{1}{5}}{2 x - 1} + \frac{- \frac{1}{10}}{x + 2}$

Step 1.5

Simplify.

Tap for more steps...

Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{2} \cdot \frac{1}{x} + \frac{\frac{1}{5}}{2 x - 1} + \frac{- \frac{1}{10}}{x + 2}$

Step 1.5.2

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

$\frac{1}{2 x} + \frac{\frac{1}{5}}{2 x - 1} + \frac{- \frac{1}{10}}{x + 2}$

Step 1.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{2 x} + \frac{1}{5} \cdot \frac{1}{2 x - 1} + \frac{- \frac{1}{10}}{x + 2}$

Step 1.5.4

Multiply $\frac{1}{5}$ by $\frac{1}{2 x - 1}$ .

$\frac{1}{2 x} + \frac{1}{5 (2 x - 1)} + \frac{- \frac{1}{10}}{x + 2}$

Step 1.5.5

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{2 x} + \frac{1}{5 (2 x - 1)} - \frac{1}{10} \cdot \frac{1}{x + 2}$

Step 1.5.6

Multiply $\frac{1}{x + 2}$ by $\frac{1}{10}$ .

$\frac{1}{2 x} + \frac{1}{5 (2 x - 1)} - \frac{1}{(x + 2) \cdot 10}$

Step 1.5.7

Move $10$ to the left of $x + 2$ .

$\int \frac{1}{2 x} + \frac{1}{5 (2 x - 1)} - \frac{1}{10 (x + 2)} d x$

Step 2

Split the single integral into multiple integrals.

$\int \frac{1}{2 x} d x + \int \frac{1}{5 (2 x - 1)} d x + \int - \frac{1}{10 (x + 2)} d x$

Step 3

Since $\frac{1}{2}$ is constant with respect to $x$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} \int \frac{1}{x} d x + \int \frac{1}{5 (2 x - 1)} d x + \int - \frac{1}{10 (x + 2)} d x$

Step 4

The integral of $\frac{1}{x}$ with respect to $x$ is $\ln (| x |)$ .

$\frac{1}{2} (\ln (| x |) + C) + \int \frac{1}{5 (2 x - 1)} d x + \int - \frac{1}{10 (x + 2)} d x$

Step 5

Since $\frac{1}{5}$ is constant with respect to $x$ , move $\frac{1}{5}$ out of the integral.

$\frac{1}{2} (\ln (| x |) + C) + \frac{1}{5} \int \frac{1}{2 x - 1} d x + \int - \frac{1}{10 (x + 2)} d x$

Step 6

Let $u_{1} = 2 x - 1$ . Then $d u_{1} = 2 d x$ , so $\frac{1}{2} d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

Tap for more steps...

Step 6.1

Let $u_{1} = 2 x - 1$ . Find $\frac{d u_{1}}{d x}$ .

Tap for more steps...

Step 6.1.1

Differentiate $2 x - 1$ .

$\frac{d}{d x} [2 x - 1]$

Step 6.1.2

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 6.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

Tap for more steps...

Step 6.1.3.1

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 6.1.3.2

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 6.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [- 1]$

Step 6.1.4

Differentiate using the Constant Rule.

Tap for more steps...

Step 6.1.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$2 + 0$

Step 6.1.4.2

Add $2$ and $0$ .

$2$

Step 6.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\frac{1}{2} (\ln (| x |) + C) + \frac{1}{5} \int \frac{1}{u_{1}} \cdot \frac{1}{2} d u_{1} + \int - \frac{1}{10 (x + 2)} d x$

Step 7

Simplify.

Tap for more steps...

Step 7.1

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

$\frac{1}{2} (\ln (| x |) + C) + \frac{1}{5} \int \frac{1}{u_{1} \cdot 2} d u_{1} + \int - \frac{1}{10 (x + 2)} d x$

Step 7.2

Move $2$ to the left of $u_{1}$ .

$\frac{1}{2} (\ln (| x |) + C) + \frac{1}{5} \int \frac{1}{2 u_{1}} d u_{1} + \int - \frac{1}{10 (x + 2)} d x$

Step 8

Since $\frac{1}{2}$ is constant with respect to $u_{1}$ , move $\frac{1}{2}$ out of the integral.

$\frac{1}{2} (\ln (| x |) + C) + \frac{1}{5} (\frac{1}{2} \int \frac{1}{u_{1}} d u_{1}) + \int - \frac{1}{10 (x + 2)} d x$

Step 9

Simplify.

Tap for more steps...

Step 9.1

Multiply $\frac{1}{2}$ by $\frac{1}{5}$ .

$\frac{1}{2} (\ln (| x |) + C) + \frac{1}{2 \cdot 5} \int \frac{1}{u_{1}} d u_{1} + \int - \frac{1}{10 (x + 2)} d x$

Step 9.2

Multiply $2$ by $5$ .

$\frac{1}{2} (\ln (| x |) + C) + \frac{1}{10} \int \frac{1}{u_{1}} d u_{1} + \int - \frac{1}{10 (x + 2)} d x$

Step 10

The integral of $\frac{1}{u_{1}}$ with respect to $u_{1}$ is $\ln (| u_{1} |)$ .

$\frac{1}{2} (\ln (| x |) + C) + \frac{1}{10} (\ln (| u_{1} |) + C) + \int - \frac{1}{10 (x + 2)} d x$

Step 11

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$\frac{1}{2} (\ln (| x |) + C) + \frac{1}{10} (\ln (| u_{1} |) + C) - \int \frac{1}{10 (x + 2)} d x$

Step 12

Since $\frac{1}{10}$ is constant with respect to $x$ , move $\frac{1}{10}$ out of the integral.

$\frac{1}{2} (\ln (| x |) + C) + \frac{1}{10} (\ln (| u_{1} |) + C) - (\frac{1}{10} \int \frac{1}{x + 2} d x)$

Step 13

Let $u_{2} = x + 2$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

Tap for more steps...

Step 13.1

Let $u_{2} = x + 2$ . Find $\frac{d u_{2}}{d x}$ .

Tap for more steps...

Step 13.1.1

Differentiate $x + 2$ .

$\frac{d}{d x} [x + 2]$

Step 13.1.2

By the Sum Rule, the derivative of $x + 2$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [2]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [2]$

Step 13.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$1 + \frac{d}{d x} [2]$

Step 13.1.4

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$1 + 0$

Step 13.1.5

Add $1$ and $0$ .

$1$

Step 13.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\frac{1}{2} (\ln (| x |) + C) + \frac{1}{10} (\ln (| u_{1} |) + C) - \frac{1}{10} \int \frac{1}{u_{2}} d u_{2}$

Step 14

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$\frac{1}{2} (\ln (| x |) + C) + \frac{1}{10} (\ln (| u_{1} |) + C) - \frac{1}{10} (\ln (| u_{2} |) + C)$

Step 15

Simplify.

$\frac{1}{2} \ln (| x |) + \frac{1}{10} \ln (| u_{1} |) - \frac{1}{10} \ln (| u_{2} |) + C$

Step 16

Substitute back in for each integration substitution variable.

Tap for more steps...

Step 16.1

Replace all occurrences of $u_{1}$ with $2 x - 1$ .

$\frac{1}{2} \ln (| x |) + \frac{1}{10} \ln (| 2 x - 1 |) - \frac{1}{10} \ln (| u_{2} |) + C$

Step 16.2

Replace all occurrences of $u_{2}$ with $x + 2$ .

$\frac{1}{2} \ln (| x |) + \frac{1}{10} \ln (| 2 x - 1 |) - \frac{1}{10} \ln (| x + 2 |) + C$

Enter YOUR Problem