Calculus Examples

$\frac{5}{{(2 - 8 x)}^{3}}$

Step 1

Write $\frac{5}{{(2 - 8 x)}^{3}}$ as a function.

$f (x) = \frac{5}{{(2 - 8 x)}^{3}}$

Step 2

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

$F (x) = \int \frac{5}{{(2 - 8 x)}^{3}} d x$

Step 4

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$5 \int \frac{1}{{(2 - 8 x)}^{3}} d x$

Step 5

Write the fraction using partial fraction decomposition.

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Step 5.1

Decompose the fraction and multiply through by the common denominator.

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Step 5.1.1

Factor the fraction.

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Step 5.1.1.1

Factor $2$ out of $2 - 8 x$ .

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Step 5.1.1.1.1

Factor $2$ out of $2$ .

$\frac{1}{{(2 (1) - 8 x)}^{3}}$

Step 5.1.1.1.2

Factor $2$ out of $- 8 x$ .

$\frac{1}{{(2 (1) + 2 (- 4 x))}^{3}}$

Step 5.1.1.1.3

Factor $2$ out of $2 (1) + 2 (- 4 x)$ .

$\frac{1}{{(2 (1 - 4 x))}^{3}}$

Step 5.1.1.2

Apply the product rule to $2 (1 - 4 x)$ .

$\frac{1}{2^{3} {(1 - 4 x)}^{3}}$

Step 5.1.2

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 $A$ .

$\frac{A}{{(1 - 4 x)}^{3}}$

Step 5.1.3

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

Step 5.1.4

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

Step 5.1.5

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $2^{3} {(1 - 4 x)}^{3}$ .

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

Step 5.1.6

Cancel the common factor of $2^{3}$ .

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Step 5.1.6.1

Cancel the common factor.

Step 5.1.6.2

Rewrite the expression.

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

Step 5.1.7

Cancel the common factor of ${(1 - 4 x)}^{3}$ .

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Step 5.1.7.1

Cancel the common factor.

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

Step 5.1.7.2

Rewrite the expression.

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

Step 5.1.8

Simplify each term.

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Step 5.1.8.1

Cancel the common factor of ${(1 - 4 x)}^{3}$ .

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Step 5.1.8.1.1

Cancel the common factor.

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

Step 5.1.8.1.2

Divide $(A) \cdot (2^{3})$ by $1$ .

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

Step 5.1.8.2

Raise $2$ to the power of $3$ .

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

Step 5.1.8.3

Move $8$ to the left of $A$ .

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

Step 5.1.8.4

Cancel the common factor of ${(1 - 4 x)}^{3}$ and ${(1 - 4 x)}^{2}$ .

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Step 5.1.8.4.1

Factor ${(1 - 4 x)}^{2}$ out of $(B) (2^{3} {(1 - 4 x)}^{3})$ .

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

Step 5.1.8.4.2

Cancel the common factors.

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Step 5.1.8.4.2.1

Multiply by $1$ .

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

Step 5.1.8.4.2.2

Cancel the common factor.

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

Step 5.1.8.4.2.3

Rewrite the expression.

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

Step 5.1.8.4.2.4

Divide $B (2^{3} (1 - 4 x))$ by $1$ .

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

Step 5.1.8.5

Rewrite using the commutative property of multiplication.

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

Step 5.1.8.6

Raise $2$ to the power of $3$ .

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

Step 5.1.8.7

Apply the distributive property.

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

Step 5.1.8.8

Multiply $8$ by $1$ .

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

Step 5.1.8.9

Rewrite using the commutative property of multiplication.

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

Step 5.1.8.10

Multiply $8$ by $- 4$ .

$1 = 8 A + 8 B - 32 B x + \frac{(C) (2^{3} {(1 - 4 x)}^{3})}{1 - 4 x}$

Step 5.1.8.11

Cancel the common factor of ${(1 - 4 x)}^{3}$ and $1 - 4 x$ .

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Step 5.1.8.11.1

Factor $1 - 4 x$ out of $(C) (2^{3} {(1 - 4 x)}^{3})$ .

$1 = 8 A + 8 B - 32 B x + \frac{(1 - 4 x) (C (2^{3} {(1 - 4 x)}^{2}))}{1 - 4 x}$

Step 5.1.8.11.2

Cancel the common factors.

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Step 5.1.8.11.2.1

Multiply by $1$ .

$1 = 8 A + 8 B - 32 B x + \frac{(1 - 4 x) (C (2^{3} {(1 - 4 x)}^{2}))}{(1 - 4 x) \cdot 1}$

Step 5.1.8.11.2.2

Cancel the common factor.

$1 = 8 A + 8 B - 32 B x + \frac{(1 - 4 x) (C (2^{3} {(1 - 4 x)}^{2}))}{(1 - 4 x) \cdot 1}$

Step 5.1.8.11.2.3

Rewrite the expression.

$1 = 8 A + 8 B - 32 B x + \frac{C (2^{3} {(1 - 4 x)}^{2})}{1}$

Step 5.1.8.11.2.4

Divide $C (2^{3} {(1 - 4 x)}^{2})$ by $1$ .

$1 = 8 A + 8 B - 32 B x + C (2^{3} {(1 - 4 x)}^{2})$

Step 5.1.8.12

Rewrite using the commutative property of multiplication.

$1 = 8 A + 8 B - 32 B x + 2^{3} C {(1 - 4 x)}^{2}$

Step 5.1.8.13

Raise $2$ to the power of $3$ .

$1 = 8 A + 8 B - 32 B x + 8 C {(1 - 4 x)}^{2}$

Step 5.1.8.14

Rewrite ${(1 - 4 x)}^{2}$ as $(1 - 4 x) (1 - 4 x)$ .

$1 = 8 A + 8 B - 32 B x + 8 C ((1 - 4 x) (1 - 4 x))$

Step 5.1.8.15

Expand $(1 - 4 x) (1 - 4 x)$ using the FOIL Method.

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Step 5.1.8.15.1

Apply the distributive property.

$1 = 8 A + 8 B - 32 B x + 8 C (1 (1 - 4 x) - 4 x (1 - 4 x))$

Step 5.1.8.15.2

Apply the distributive property.

$1 = 8 A + 8 B - 32 B x + 8 C (1 \cdot 1 + 1 (- 4 x) - 4 x (1 - 4 x))$

Step 5.1.8.15.3

Apply the distributive property.

$1 = 8 A + 8 B - 32 B x + 8 C (1 \cdot 1 + 1 (- 4 x) - 4 x \cdot 1 - 4 x (- 4 x))$

Step 5.1.8.16

Simplify and combine like terms.

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Step 5.1.8.16.1

Simplify each term.

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Step 5.1.8.16.1.1

Multiply $1$ by $1$ .

$1 = 8 A + 8 B - 32 B x + 8 C (1 + 1 (- 4 x) - 4 x \cdot 1 - 4 x (- 4 x))$

Step 5.1.8.16.1.2

Multiply $- 4 x$ by $1$ .

$1 = 8 A + 8 B - 32 B x + 8 C (1 - 4 x - 4 x \cdot 1 - 4 x (- 4 x))$

Step 5.1.8.16.1.3

Multiply $- 4$ by $1$ .

$1 = 8 A + 8 B - 32 B x + 8 C (1 - 4 x - 4 x - 4 x (- 4 x))$

Step 5.1.8.16.1.4

Rewrite using the commutative property of multiplication.

$1 = 8 A + 8 B - 32 B x + 8 C (1 - 4 x - 4 x - 4 \cdot (- 4 x \cdot x))$

Step 5.1.8.16.1.5

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

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Step 5.1.8.16.1.5.1

Move $x$ .

$1 = 8 A + 8 B - 32 B x + 8 C (1 - 4 x - 4 x - 4 \cdot (- 4 (x \cdot x)))$

Step 5.1.8.16.1.5.2

Multiply $x$ by $x$ .

$1 = 8 A + 8 B - 32 B x + 8 C (1 - 4 x - 4 x - 4 \cdot (- 4 x^{2}))$

Step 5.1.8.16.1.6

Multiply $- 4$ by $- 4$ .

$1 = 8 A + 8 B - 32 B x + 8 C (1 - 4 x - 4 x + 16 x^{2})$

Step 5.1.8.16.2

Subtract $4 x$ from $- 4 x$ .

$1 = 8 A + 8 B - 32 B x + 8 C (1 - 8 x + 16 x^{2})$

Step 5.1.8.17

Apply the distributive property.

$1 = 8 A + 8 B - 32 B x + 8 C \cdot 1 + 8 C (- 8 x) + 8 C (16 x^{2})$

Step 5.1.8.18

Simplify.

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Step 5.1.8.18.1

Multiply $8$ by $1$ .

$1 = 8 A + 8 B - 32 B x + 8 C + 8 C (- 8 x) + 8 C (16 x^{2})$

Step 5.1.8.18.2

Rewrite using the commutative property of multiplication.

$1 = 8 A + 8 B - 32 B x + 8 C + 8 \cdot (- 8 C x) + 8 C (16 x^{2})$

Step 5.1.8.18.3

Rewrite using the commutative property of multiplication.

$1 = 8 A + 8 B - 32 B x + 8 C + 8 \cdot (- 8 C x) + 8 \cdot (16 C x^{2})$

Step 5.1.8.19

Simplify each term.

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Step 5.1.8.19.1

Multiply $8$ by $- 8$ .

$1 = 8 A + 8 B - 32 B x + 8 C - 64 C x + 8 \cdot (16 C x^{2})$

Step 5.1.8.19.2

Multiply $8$ by $16$ .

$1 = 8 A + 8 B - 32 B x + 8 C - 64 C x + 128 C x^{2}$

Step 5.1.9

Simplify the expression.

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Step 5.1.9.1

Move $B$ .

$1 = 8 A + 8 B - 32 x B + 8 C - 64 C x + 128 C x^{2}$

Step 5.1.9.2

Move $8 C$ .

$1 = 8 A + 8 B - 32 x B - 64 C x + 128 C x^{2} + 8 C$

Step 5.1.9.3

Move $8 B$ .

$1 = 8 A - 32 x B - 64 C x + 128 C x^{2} + 8 B + 8 C$

Step 5.1.9.4

Move $8 A$ .

$1 = - 32 x B - 64 C x + 128 C x^{2} + 8 A + 8 B + 8 C$

Step 5.1.9.5

Move $- 64 C x$ .

$1 = - 32 x B + 128 C x^{2} - 64 C x + 8 A + 8 B + 8 C$

Step 5.1.9.6

Reorder $- 32 x B$ and $128 C x^{2}$ .

$1 = 128 C x^{2} - 32 x B - 64 C x + 8 A + 8 B + 8 C$

Step 5.2

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

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Step 5.2.1

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.

$0 = 128 C$

Step 5.2.2

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.

$0 = - 32 B - 64 C$

Step 5.2.3

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.

$1 = 8 A + 8 B + 8 C$

Step 5.2.4

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

$0 = 128 C$

$0 = - 32 B - 64 C$

$1 = 8 A + 8 B + 8 C$

$0 = 128 C$

$0 = - 32 B - 64 C$

$1 = 8 A + 8 B + 8 C$

Step 5.3

Solve the system of equations.

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Step 5.3.1

Solve for $C$ in $0 = 128 C$ .

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Step 5.3.1.1

Rewrite the equation as $128 C = 0$ .

$128 C = 0$

$0 = - 32 B - 64 C$

$1 = 8 A + 8 B + 8 C$

Step 5.3.1.2

Divide each term in $128 C = 0$ by $128$ and simplify.

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Step 5.3.1.2.1

Divide each term in $128 C = 0$ by $128$ .

$\frac{128 C}{128} = \frac{0}{128}$

$0 = - 32 B - 64 C$

$1 = 8 A + 8 B + 8 C$

Step 5.3.1.2.2

Simplify the left side.

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Step 5.3.1.2.2.1

Cancel the common factor of $128$ .

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Step 5.3.1.2.2.1.1

Cancel the common factor.

$\frac{128 C}{128} = \frac{0}{128}$

$0 = - 32 B - 64 C$

$1 = 8 A + 8 B + 8 C$

Step 5.3.1.2.2.1.2

Divide $C$ by $1$ .

$C = \frac{0}{128}$

$0 = - 32 B - 64 C$

$1 = 8 A + 8 B + 8 C$

$C = \frac{0}{128}$

$0 = - 32 B - 64 C$

$1 = 8 A + 8 B + 8 C$

$C = \frac{0}{128}$

$0 = - 32 B - 64 C$

$1 = 8 A + 8 B + 8 C$

Step 5.3.1.2.3

Simplify the right side.

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Step 5.3.1.2.3.1

Divide $0$ by $128$ .

$C = 0$

$0 = - 32 B - 64 C$

$1 = 8 A + 8 B + 8 C$

$C = 0$

$0 = - 32 B - 64 C$

$1 = 8 A + 8 B + 8 C$

$C = 0$

$0 = - 32 B - 64 C$

$1 = 8 A + 8 B + 8 C$

$C = 0$

$0 = - 32 B - 64 C$

$1 = 8 A + 8 B + 8 C$

Step 5.3.2

Replace all occurrences of $C$ with $0$ in each equation.

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Step 5.3.2.1

Replace all occurrences of $C$ in $0 = - 32 B - 64 C$ with $0$ .

$0 = - 32 B - 64 \cdot 0$

$C = 0$

$1 = 8 A + 8 B + 8 C$

Step 5.3.2.2

Simplify the right side.

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Step 5.3.2.2.1

Simplify $- 32 B - 64 \cdot 0$ .

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Step 5.3.2.2.1.1

Multiply $- 64$ by $0$ .

$0 = - 32 B + 0$

$C = 0$

$1 = 8 A + 8 B + 8 C$

Step 5.3.2.2.1.2

Add $- 32 B$ and $0$ .

$0 = - 32 B$

$C = 0$

$1 = 8 A + 8 B + 8 C$

$0 = - 32 B$

$C = 0$

$1 = 8 A + 8 B + 8 C$

$0 = - 32 B$

$C = 0$

$1 = 8 A + 8 B + 8 C$

Step 5.3.2.3

Replace all occurrences of $C$ in $1 = 8 A + 8 B + 8 C$ with $0$ .

$1 = 8 A + 8 B + 8 (0)$

$0 = - 32 B$

$C = 0$

Step 5.3.2.4

Simplify the right side.

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Step 5.3.2.4.1

Simplify $8 A + 8 B + 8 (0)$ .

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Step 5.3.2.4.1.1

Multiply $8$ by $0$ .

$1 = 8 A + 8 B + 0$

$0 = - 32 B$

$C = 0$

Step 5.3.2.4.1.2

Add $8 A + 8 B$ and $0$ .

$1 = 8 A + 8 B$

$0 = - 32 B$

$C = 0$

$1 = 8 A + 8 B$

$0 = - 32 B$

$C = 0$

$1 = 8 A + 8 B$

$0 = - 32 B$

$C = 0$

$1 = 8 A + 8 B$

$0 = - 32 B$

$C = 0$

Step 5.3.3

Solve for $B$ in $0 = - 32 B$ .

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Step 5.3.3.1

Rewrite the equation as $- 32 B = 0$ .

$- 32 B = 0$

$1 = 8 A + 8 B$

$C = 0$

Step 5.3.3.2

Divide each term in $- 32 B = 0$ by $- 32$ and simplify.

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Step 5.3.3.2.1

Divide each term in $- 32 B = 0$ by $- 32$ .

$\frac{- 32 B}{- 32} = \frac{0}{- 32}$

$1 = 8 A + 8 B$

$C = 0$

Step 5.3.3.2.2

Simplify the left side.

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Step 5.3.3.2.2.1

Cancel the common factor of $- 32$ .

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Step 5.3.3.2.2.1.1

Cancel the common factor.

$\frac{- 32 B}{- 32} = \frac{0}{- 32}$

$1 = 8 A + 8 B$

$C = 0$

Step 5.3.3.2.2.1.2

Divide $B$ by $1$ .

$B = \frac{0}{- 32}$

$1 = 8 A + 8 B$

$C = 0$

$B = \frac{0}{- 32}$

$1 = 8 A + 8 B$

$C = 0$

$B = \frac{0}{- 32}$

$1 = 8 A + 8 B$

$C = 0$

Step 5.3.3.2.3

Simplify the right side.

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Step 5.3.3.2.3.1

Divide $0$ by $- 32$ .

$B = 0$

$1 = 8 A + 8 B$

$C = 0$

$B = 0$

$1 = 8 A + 8 B$

$C = 0$

$B = 0$

$1 = 8 A + 8 B$

$C = 0$

$B = 0$

$1 = 8 A + 8 B$

$C = 0$

Step 5.3.4

Replace all occurrences of $B$ with $0$ in each equation.

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Step 5.3.4.1

Replace all occurrences of $B$ in $1 = 8 A + 8 B$ with $0$ .

$1 = 8 A + 8 (0)$

$B = 0$

$C = 0$

Step 5.3.4.2

Simplify the right side.

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Step 5.3.4.2.1

Simplify $8 A + 8 (0)$ .

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Step 5.3.4.2.1.1

Multiply $8$ by $0$ .

$1 = 8 A + 0$

$B = 0$

$C = 0$

Step 5.3.4.2.1.2

Add $8 A$ and $0$ .

$1 = 8 A$

$B = 0$

$C = 0$

$1 = 8 A$

$B = 0$

$C = 0$

$1 = 8 A$

$B = 0$

$C = 0$

$1 = 8 A$

$B = 0$

$C = 0$

Step 5.3.5

Solve for $A$ in $1 = 8 A$ .

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Step 5.3.5.1

Rewrite the equation as $8 A = 1$ .

$8 A = 1$

$B = 0$

$C = 0$

Step 5.3.5.2

Divide each term in $8 A = 1$ by $8$ and simplify.

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Step 5.3.5.2.1

Divide each term in $8 A = 1$ by $8$ .

$\frac{8 A}{8} = \frac{1}{8}$

$B = 0$

$C = 0$

Step 5.3.5.2.2

Simplify the left side.

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Step 5.3.5.2.2.1

Cancel the common factor of $8$ .

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Step 5.3.5.2.2.1.1

Cancel the common factor.

$\frac{8 A}{8} = \frac{1}{8}$

$B = 0$

$C = 0$

Step 5.3.5.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{1}{8}$

$B = 0$

$C = 0$

$A = \frac{1}{8}$

$B = 0$

$C = 0$

$A = \frac{1}{8}$

$B = 0$

$C = 0$

$A = \frac{1}{8}$

$B = 0$

$C = 0$

$A = \frac{1}{8}$

$B = 0$

$C = 0$

Step 5.3.6

Solve the system of equations.

$A = \frac{1}{8} B = 0 C = 0$

Step 5.3.7

List all of the solutions.

$A = \frac{1}{8}, B = 0, C = 0$

Step 5.4

Replace each of the partial fraction coefficients in $\frac{A}{{(1 - 4 x)}^{3}} + \frac{B}{{(1 - 4 x)}^{2}} + \frac{C}{1 - 4 x}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{\frac{1}{8}}{{(1 - 4 x)}^{3}} + \frac{0}{{(1 - 4 x)}^{2}} + \frac{0}{1 - 4 x}$

Step 5.5

Simplify.

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Step 5.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{8} \cdot \frac{1}{{(1 - 4 x)}^{3}} + \frac{0}{{(1 - 4 x)}^{2}} + \frac{0}{1 - 4 x}$

Step 5.5.2

Combine.

$\frac{1 \cdot 1}{8 {(1 - 4 x)}^{3}} + \frac{0}{{(1 - 4 x)}^{2}} + \frac{0}{1 - 4 x}$

Step 5.5.3

Multiply $1$ by $1$ .

$\frac{1}{8 {(1 - 4 x)}^{3}} + \frac{0}{{(1 - 4 x)}^{2}} + \frac{0}{1 - 4 x}$

Step 5.5.4

Divide $0$ by ${(1 - 4 x)}^{2}$ .

$\frac{1}{8 {(1 - 4 x)}^{3}} + 0 + \frac{0}{1 - 4 x}$

Step 5.5.5

Divide $0$ by $1 - 4 x$ .

$\frac{1}{8 {(1 - 4 x)}^{3}} + 0 + 0$

Step 5.5.6

Remove the zero from the expression.

$5 \int \frac{1}{8 {(1 - 4 x)}^{3}} d x$

Step 6

Since $\frac{1}{8}$ is constant with respect to $x$ , move $\frac{1}{8}$ out of the integral.

$5 (\frac{1}{8} \int \frac{1}{{(1 - 4 x)}^{3}} d x)$

Step 7

Combine $\frac{1}{8}$ and $5$ .

$\frac{5}{8} \int \frac{1}{{(1 - 4 x)}^{3}} d x$

Step 8

Let $u = 1 - 4 x$ . Then $d u = - 4 d x$ , so $- \frac{1}{4} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 8.1

Let $u = 1 - 4 x$ . Find $\frac{d u}{d x}$ .

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Step 8.1.1

Differentiate $1 - 4 x$ .

$\frac{d}{d x} [1 - 4 x]$

Step 8.1.2

Differentiate.

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Step 8.1.2.1

By the Sum Rule, the derivative of $1 - 4 x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 4 x]$ .

$\frac{d}{d x} [1] + \frac{d}{d x} [- 4 x]$

Step 8.1.2.2

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

$0 + \frac{d}{d x} [- 4 x]$

Step 8.1.3

Evaluate $\frac{d}{d x} [- 4 x]$ .

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Step 8.1.3.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$0 - 4 \frac{d}{d x} [x]$

Step 8.1.3.2

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

$0 - 4 \cdot 1$

Step 8.1.3.3

Multiply $- 4$ by $1$ .

$0 - 4$

Step 8.1.4

Subtract $4$ from $0$ .

$- 4$

Step 8.2

Rewrite the problem using $u$ and $d u$ .

$\frac{5}{8} \int \frac{1}{u^{3}} \cdot \frac{1}{- 4} d u$

Step 9

Simplify.

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Step 9.1

Move the negative in front of the fraction.

$\frac{5}{8} \int \frac{1}{u^{3}} (- \frac{1}{4}) d u$

Step 9.2

Multiply $\frac{1}{u^{3}}$ by $\frac{1}{4}$ .

$\frac{5}{8} \int - \frac{1}{u^{3} \cdot 4} d u$

Step 9.3

Move $4$ to the left of $u^{3}$ .

$\frac{5}{8} \int - \frac{1}{4 u^{3}} d u$

Step 10

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

$\frac{5}{8} (- \int \frac{1}{4 u^{3}} d u)$

Step 11

Since $\frac{1}{4}$ is constant with respect to $u$ , move $\frac{1}{4}$ out of the integral.

$\frac{5}{8} (- (\frac{1}{4} \int \frac{1}{u^{3}} d u))$

Step 12

Simplify the expression.

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Step 12.1

Simplify.

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Step 12.1.1

Multiply $\frac{5}{8}$ by $\frac{1}{4}$ .

$- \frac{5}{8 \cdot 4} \int \frac{1}{u^{3}} d u$

Step 12.1.2

Multiply $8$ by $4$ .

$- \frac{5}{32} \int \frac{1}{u^{3}} d u$

Step 12.2

Apply basic rules of exponents.

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Step 12.2.1

Move $u^{3}$ out of the denominator by raising it to the $- 1$ power.

$- \frac{5}{32} \int {(u^{3})}^{- 1} d u$

Step 12.2.2

Multiply the exponents in ${(u^{3})}^{- 1}$ .

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Step 12.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- \frac{5}{32} \int u^{3 \cdot - 1} d u$

Step 12.2.2.2

Multiply $3$ by $- 1$ .

$- \frac{5}{32} \int u^{- 3} d u$

Step 13

By the Power Rule, the integral of $u^{- 3}$ with respect to $u$ is $- \frac{1}{2} u^{- 2}$ .

$- \frac{5}{32} (- \frac{1}{2} u^{- 2} + C)$

Step 14

Simplify.

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Step 14.1

Rewrite $- \frac{5}{32} (- \frac{1}{2} u^{- 2} + C)$ as $- \frac{5}{32} (- \frac{1}{2} \cdot \frac{1}{u^{2}}) + C$ .

$- \frac{5}{32} (- \frac{1}{2} \cdot \frac{1}{u^{2}}) + C$

Step 14.2

Simplify.

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Step 14.2.1

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

$- \frac{5}{32} (- \frac{1}{u^{2} \cdot 2}) + C$

Step 14.2.2

Move $2$ to the left of $u^{2}$ .

$- \frac{5}{32} (- \frac{1}{2 \cdot u^{2}}) + C$

Step 14.2.3

Multiply $- 1$ by $- 1$ .

$1 (\frac{5}{32}) \frac{1}{2 u^{2}} + C$

Step 14.2.4

Multiply $\frac{5}{32}$ by $1$ .

$\frac{5}{32} \cdot \frac{1}{2 u^{2}} + C$

Step 14.2.5

Multiply $\frac{5}{32}$ by $\frac{1}{2 u^{2}}$ .

$\frac{5}{32 (2 u^{2})} + C$

Step 14.2.6

Multiply $2$ by $32$ .

$\frac{5}{64 u^{2}} + C$

Step 15

Replace all occurrences of $u$ with $1 - 4 x$ .

$\frac{5}{64 {(1 - 4 x)}^{2}} + C$

Step 16

The answer is the antiderivative of the function $f (x) = \frac{5}{{(2 - 8 x)}^{3}}$ .

$F (x) =$ $\frac{5}{64 {(1 - 4 x)}^{2}} + C$