Calculus Examples

$\frac{1}{x^{3} - x}$

Step 1

Write $\frac{1}{x^{3} - x}$ as a function.

$f (x) = \frac{1}{x^{3} - x}$

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.

$F (x) = \int \frac{1}{x^{3} - x} d x$

Step 4

Write the fraction using partial fraction decomposition.

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Step 4.1

Decompose the fraction and multiply through by the common denominator.

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Step 4.1.1

Factor the fraction.

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Step 4.1.1.1

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

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Step 4.1.1.1.1

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

$\frac{1}{x \cdot x^{2} - x}$

Step 4.1.1.1.2

Factor $x$ out of $- x$ .

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

Step 4.1.1.1.3

Factor $x$ out of $x \cdot x^{2} + x \cdot - 1$ .

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

Step 4.1.1.2

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

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

Step 4.1.1.3

Factor.

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Step 4.1.1.3.1

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

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

Step 4.1.1.3.2

Remove unnecessary parentheses.

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

Step 4.1.2

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 $B$ .

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

Step 4.1.3

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

Step 4.1.4

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

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

Step 4.1.5

Cancel the common factor of $x$ .

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Step 4.1.5.1

Cancel the common factor.

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

Step 4.1.5.2

Rewrite the expression.

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

Step 4.1.6

Cancel the common factor of $x + 1$ .

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Step 4.1.6.1

Cancel the common factor.

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

Step 4.1.6.2

Rewrite the expression.

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

Step 4.1.7

Cancel the common factor of $x - 1$ .

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Step 4.1.7.1

Cancel the common factor.

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

Step 4.1.7.2

Rewrite the expression.

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

Step 4.1.8

Simplify each term.

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Step 4.1.8.1

Cancel the common factor of $x$ .

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Step 4.1.8.1.1

Cancel the common factor.

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

Step 4.1.8.1.2

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

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

Step 4.1.8.2

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

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Step 4.1.8.2.1

Apply the distributive property.

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

Step 4.1.8.2.2

Apply the distributive property.

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

Step 4.1.8.2.3

Apply the distributive property.

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

Step 4.1.8.3

Simplify and combine like terms.

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Step 4.1.8.3.1

Simplify each term.

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Step 4.1.8.3.1.1

Multiply $x$ by $x$ .

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

Step 4.1.8.3.1.2

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

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

Step 4.1.8.3.1.3

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

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

Step 4.1.8.3.1.4

Multiply $x$ by $1$ .

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

Step 4.1.8.3.1.5

Multiply $- 1$ by $1$ .

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

Step 4.1.8.3.2

Add $- x$ and $x$ .

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

Step 4.1.8.3.3

Add $x^{2}$ and $0$ .

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

Step 4.1.8.4

Apply the distributive property.

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

Step 4.1.8.5

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

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

Step 4.1.8.6

Rewrite $- 1 A$ as $- A$ .

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

Step 4.1.8.7

Cancel the common factor of $x + 1$ .

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Step 4.1.8.7.1

Cancel the common factor.

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

Step 4.1.8.7.2

Divide $(B) (x (x - 1))$ by $1$ .

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

Step 4.1.8.8

Apply the distributive property.

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

Step 4.1.8.9

Multiply $x$ by $x$ .

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

Step 4.1.8.10

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

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

Step 4.1.8.11

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

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

Step 4.1.8.12

Apply the distributive property.

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

Step 4.1.8.13

Rewrite using the commutative property of multiplication.

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

Step 4.1.8.14

Cancel the common factor of $x - 1$ .

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Step 4.1.8.14.1

Cancel the common factor.

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

Step 4.1.8.14.2

Divide $(C) (x (x + 1))$ by $1$ .

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

Step 4.1.8.15

Apply the distributive property.

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

Step 4.1.8.16

Multiply $x$ by $x$ .

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

Step 4.1.8.17

Multiply $x$ by $1$ .

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

Step 4.1.8.18

Apply the distributive property.

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

Step 4.1.9

Simplify the expression.

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Step 4.1.9.1

Move $B$ .

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

Step 4.1.9.2

Move $- A$ .

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

Step 4.1.9.3

Move $- x B$ .

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

Step 4.2

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

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Step 4.2.1

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.

$0 = A + B + C$

Step 4.2.2

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.

$0 = - 1 B + C$

Step 4.2.3

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.

$1 = - 1 A$

Step 4.2.4

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

$0 = A + B + C$

$0 = - 1 B + C$

$1 = - 1 A$

$0 = A + B + C$

$0 = - 1 B + C$

$1 = - 1 A$

Step 4.3

Solve the system of equations.

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Step 4.3.1

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

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Step 4.3.1.1

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

$- 1 A = 1$

$0 = A + B + C$

$0 = - 1 B + C$

Step 4.3.1.2

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

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Step 4.3.1.2.1

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

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

$0 = A + B + C$

$0 = - 1 B + C$

Step 4.3.1.2.2

Simplify the left side.

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Step 4.3.1.2.2.1

Dividing two negative values results in a positive value.

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

$0 = A + B + C$

$0 = - 1 B + C$

Step 4.3.1.2.2.2

Divide $A$ by $1$ .

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

$0 = A + B + C$

$0 = - 1 B + C$

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

$0 = A + B + C$

$0 = - 1 B + C$

Step 4.3.1.2.3

Simplify the right side.

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Step 4.3.1.2.3.1

Divide $1$ by $- 1$ .

$A = - 1$

$0 = A + B + C$

$0 = - 1 B + C$

$A = - 1$

$0 = A + B + C$

$0 = - 1 B + C$

$A = - 1$

$0 = A + B + C$

$0 = - 1 B + C$

$A = - 1$

$0 = A + B + C$

$0 = - 1 B + C$

Step 4.3.2

Replace all occurrences of $A$ with $- 1$ in each equation.

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Step 4.3.2.1

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

$0 = (- 1) + B + C$

$A = - 1$

$0 = - 1 B + C$

Step 4.3.2.2

Simplify the right side.

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Step 4.3.2.2.1

Remove parentheses.

$0 = - 1 + B + C$

$A = - 1$

$0 = - 1 B + C$

$0 = - 1 + B + C$

$A = - 1$

$0 = - 1 B + C$

$0 = - 1 + B + C$

$A = - 1$

$0 = - 1 B + C$

Step 4.3.3

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

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Step 4.3.3.1

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

$- 1 + B + C = 0$

$A = - 1$

$0 = - 1 B + C$

Step 4.3.3.2

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

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Step 4.3.3.2.1

Add $1$ to both sides of the equation.

$B + C = 1$

$A = - 1$

$0 = - 1 B + C$

Step 4.3.3.2.2

Subtract $C$ from both sides of the equation.

$B = 1 - C$

$A = - 1$

$0 = - 1 B + C$

$B = 1 - C$

$A = - 1$

$0 = - 1 B + C$

$B = 1 - C$

$A = - 1$

$0 = - 1 B + C$

Step 4.3.4

Replace all occurrences of $B$ with $1 - C$ in each equation.

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Step 4.3.4.1

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

$0 = - 1 (1 - C) + C$

$B = 1 - C$

$A = - 1$

Step 4.3.4.2

Simplify the right side.

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Step 4.3.4.2.1

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

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Step 4.3.4.2.1.1

Simplify each term.

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Step 4.3.4.2.1.1.1

Apply the distributive property.

$0 = - 1 \cdot 1 - 1 (- C) + C$

$B = 1 - C$

$A = - 1$

Step 4.3.4.2.1.1.2

Multiply $- 1$ by $1$ .

$0 = - 1 - 1 (- C) + C$

$B = 1 - C$

$A = - 1$

Step 4.3.4.2.1.1.3

Multiply $- 1 (- C)$ .

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Step 4.3.4.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$0 = - 1 + 1 C + C$

$B = 1 - C$

$A = - 1$

Step 4.3.4.2.1.1.3.2

Multiply $C$ by $1$ .

$0 = - 1 + C + C$

$B = 1 - C$

$A = - 1$

$0 = - 1 + C + C$

$B = 1 - C$

$A = - 1$

$0 = - 1 + C + C$

$B = 1 - C$

$A = - 1$

Step 4.3.4.2.1.2

Add $C$ and $C$ .

$0 = - 1 + 2 C$

$B = 1 - C$

$A = - 1$

$0 = - 1 + 2 C$

$B = 1 - C$

$A = - 1$

$0 = - 1 + 2 C$

$B = 1 - C$

$A = - 1$

$0 = - 1 + 2 C$

$B = 1 - C$

$A = - 1$

Step 4.3.5

Solve for $C$ in $0 = - 1 + 2 C$ .

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Step 4.3.5.1

Rewrite the equation as $- 1 + 2 C = 0$ .

$- 1 + 2 C = 0$

$B = 1 - C$

$A = - 1$

Step 4.3.5.2

Add $1$ to both sides of the equation.

$2 C = 1$

$B = 1 - C$

$A = - 1$

Step 4.3.5.3

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

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Step 4.3.5.3.1

Divide each term in $2 C = 1$ by $2$ .

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

$B = 1 - C$

$A = - 1$

Step 4.3.5.3.2

Simplify the left side.

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Step 4.3.5.3.2.1

Cancel the common factor of $2$ .

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Step 4.3.5.3.2.1.1

Cancel the common factor.

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

$B = 1 - C$

$A = - 1$

Step 4.3.5.3.2.1.2

Divide $C$ by $1$ .

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

$B = 1 - C$

$A = - 1$

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

$B = 1 - C$

$A = - 1$

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

$B = 1 - C$

$A = - 1$

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

$B = 1 - C$

$A = - 1$

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

$B = 1 - C$

$A = - 1$

Step 4.3.6

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

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Step 4.3.6.1

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

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

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

$A = - 1$

Step 4.3.6.2

Simplify the right side.

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Step 4.3.6.2.1

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

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Step 4.3.6.2.1.1

Write $1$ as a fraction with a common denominator.

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

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

$A = - 1$

Step 4.3.6.2.1.2

Combine the numerators over the common denominator.

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

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

$A = - 1$

Step 4.3.6.2.1.3

Subtract $1$ from $2$ .

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

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

$A = - 1$

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

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

$A = - 1$

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

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

$A = - 1$

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

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

$A = - 1$

Step 4.3.7

List all of the solutions.

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

Step 4.4

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

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

Step 4.5

Simplify.

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Step 4.5.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 4.5.2

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

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

Step 4.5.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 4.5.4

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

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

Step 5

Split the single integral into multiple integrals.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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

Step 9

Let $u_{1} = x + 1$ . Then $d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 9.1

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

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Step 9.1.1

Differentiate $x + 1$ .

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

Step 9.1.2

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

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

Step 9.1.3

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

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

Step 9.1.4

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

$1 + 0$

Step 9.1.5

Add $1$ and $0$ .

$1$

Step 9.2

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

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

Step 10

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

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

Step 11

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

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

Step 12

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

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Step 12.1

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

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Step 12.1.1

Differentiate $x - 1$ .

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

Step 12.1.2

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

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

Step 12.1.3

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

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

Step 12.1.4

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

$1 + 0$

Step 12.1.5

Add $1$ and $0$ .

$1$

Step 12.2

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

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

Step 13

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

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

Step 14

Simplify.

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

Step 15

Substitute back in for each integration substitution variable.

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Step 15.1

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

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

Step 15.2

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

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

Step 16

The answer is the antiderivative of the function $f (x) = \frac{1}{x^{3} - x}$ .

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