Calculus Examples

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

Step 1

Write the fraction using partial fraction decomposition.

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Step 1.1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1.1

Factor the fraction.

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Step 1.1.1.1

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

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Step 1.1.1.1.1

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

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

Step 1.1.1.1.2

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

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

Step 1.1.1.1.3

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

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

Step 1.1.1.1.4

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

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

Step 1.1.1.1.5

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

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

Step 1.1.1.2

Factor.

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Step 1.1.1.2.1

Factor $x^{2} + x - 2$ using the AC method.

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Step 1.1.1.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $1$ .

$- 1, 2$

Step 1.1.1.2.1.2

Write the factored form using these integers.

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

Step 1.1.1.2.2

Remove unnecessary parentheses.

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

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 $B$ .

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

Step 1.1.3

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

Step 1.1.4

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

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

Step 1.1.5

Cancel the common factor of $x$ .

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Step 1.1.5.1

Cancel the common factor.

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

Step 1.1.5.2

Rewrite the expression.

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

Step 1.1.6

Cancel the common factor of $x - 1$ .

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Step 1.1.6.1

Cancel the common factor.

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

Step 1.1.6.2

Rewrite the expression.

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

Step 1.1.7

Cancel the common factor of $x + 2$ .

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Step 1.1.7.1

Cancel the common factor.

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

Step 1.1.7.2

Rewrite the expression.

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

Step 1.1.8

Simplify each term.

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Step 1.1.8.1

Cancel the common factor of $x$ .

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Step 1.1.8.1.1

Cancel the common factor.

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

Step 1.1.8.1.2

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

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

Step 1.1.8.2

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

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Step 1.1.8.2.1

Apply the distributive property.

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

Step 1.1.8.2.2

Apply the distributive property.

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

Step 1.1.8.2.3

Apply the distributive property.

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

Step 1.1.8.3

Simplify and combine like terms.

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Step 1.1.8.3.1

Simplify each term.

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Step 1.1.8.3.1.1

Multiply $x$ by $x$ .

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

Step 1.1.8.3.1.2

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

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

Step 1.1.8.3.1.3

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

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

Step 1.1.8.3.1.4

Multiply $- 1$ by $2$ .

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

Step 1.1.8.3.2

Subtract $x$ from $2 x$ .

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

Step 1.1.8.4

Apply the distributive property.

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

Step 1.1.8.5

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

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

Step 1.1.8.6

Cancel the common factor of $x - 1$ .

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Step 1.1.8.6.1

Cancel the common factor.

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

Step 1.1.8.6.2

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

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

Step 1.1.8.7

Apply the distributive property.

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

Step 1.1.8.8

Multiply $x$ by $x$ .

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

Step 1.1.8.9

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

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

Step 1.1.8.10

Apply the distributive property.

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

Step 1.1.8.11

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.12

Cancel the common factor of $x + 2$ .

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Step 1.1.8.12.1

Cancel the common factor.

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

Step 1.1.8.12.2

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

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

Step 1.1.8.13

Apply the distributive property.

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

Step 1.1.8.14

Multiply $x$ by $x$ .

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

Step 1.1.8.15

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

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

Step 1.1.8.16

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

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

Step 1.1.8.17

Apply the distributive property.

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

Step 1.1.8.18

Rewrite using the commutative property of multiplication.

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

Step 1.1.9

Simplify the expression.

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Step 1.1.9.1

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

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

Step 1.1.9.2

Move $B$ .

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

Step 1.1.9.3

Move $- 2 A$ .

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

Step 1.1.9.4

Move $2 x B$ .

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

Step 1.1.9.5

Move $A x$ .

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

Step 1.2

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

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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.

$0 = A + B + C$

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.

$0 = A + 2 B - 1 C$

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.

$0 = A + B + C$

$0 = A + 2 B - 1 C$

$1 = - 2 A$

$0 = A + B + C$

$0 = A + 2 B - 1 C$

$1 = - 2 A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

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

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Step 1.3.1.1

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

$- 2 A = 1$

$0 = A + B + C$

$0 = A + 2 B - 1 C$

Step 1.3.1.2

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

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Step 1.3.1.2.1

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

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

$0 = A + B + C$

$0 = A + 2 B - 1 C$

Step 1.3.1.2.2

Simplify the left side.

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Step 1.3.1.2.2.1

Cancel the common factor of $- 2$ .

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Step 1.3.1.2.2.1.1

Cancel the common factor.

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

$0 = A + B + C$

$0 = A + 2 B - 1 C$

Step 1.3.1.2.2.1.2

Divide $A$ by $1$ .

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

$0 = A + B + C$

$0 = A + 2 B - 1 C$

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

$0 = A + B + C$

$0 = A + 2 B - 1 C$

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

$0 = A + B + C$

$0 = A + 2 B - 1 C$

Step 1.3.1.2.3

Simplify the right side.

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Step 1.3.1.2.3.1

Move the negative in front of the fraction.

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

$0 = A + B + C$

$0 = A + 2 B - 1 C$

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

$0 = A + B + C$

$0 = A + 2 B - 1 C$

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

$0 = A + B + C$

$0 = A + 2 B - 1 C$

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

$0 = A + B + C$

$0 = A + 2 B - 1 C$

Step 1.3.2

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

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Step 1.3.2.1

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

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

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

$0 = A + 2 B - 1 C$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Remove parentheses.

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

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

$0 = A + 2 B - 1 C$

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

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

$0 = A + 2 B - 1 C$

Step 1.3.2.3

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

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

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

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

Step 1.3.2.4

Simplify the right side.

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Step 1.3.2.4.1

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

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

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

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

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

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

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

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

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

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

Step 1.3.3

Solve for $B$ in $0 = - \frac{1}{2} + B + C$ .

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Step 1.3.3.1

Rewrite the equation as $- \frac{1}{2} + B + C = 0$ .

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

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

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

Step 1.3.3.2

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

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Step 1.3.3.2.1

Add $\frac{1}{2}$ to both sides of the equation.

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

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

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

Step 1.3.3.2.2

Subtract $C$ from both sides of the equation.

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

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

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

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

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

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

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

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

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

Step 1.3.4

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

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Step 1.3.4.1

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

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

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

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

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

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

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Step 1.3.4.2.1.1

Find the common denominator.

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Step 1.3.4.2.1.1.1

Write $2 (\frac{1}{2} - C)$ as a fraction with denominator $1$ .

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

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

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

Step 1.3.4.2.1.1.2

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

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

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

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

Step 1.3.4.2.1.1.3

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

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

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

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

Step 1.3.4.2.1.1.4

Write $- C$ as a fraction with denominator $1$ .

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

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

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

Step 1.3.4.2.1.1.5

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

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

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

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

Step 1.3.4.2.1.1.6

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

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

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

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

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

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

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

Step 1.3.4.2.1.2

Combine the numerators over the common denominator.

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

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

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

Step 1.3.4.2.1.3

Simplify each term.

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Step 1.3.4.2.1.3.1

Apply the distributive property.

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

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

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

Step 1.3.4.2.1.3.2

Cancel the common factor of $2$ .

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Step 1.3.4.2.1.3.2.1

Cancel the common factor.

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

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

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

Step 1.3.4.2.1.3.2.2

Rewrite the expression.

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

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

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

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

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

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

Step 1.3.4.2.1.3.3

Multiply $- 1$ by $2$ .

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

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

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

Step 1.3.4.2.1.3.4

Apply the distributive property.

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

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

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

Step 1.3.4.2.1.3.5

Multiply $2$ by $1$ .

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

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

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

Step 1.3.4.2.1.3.6

Multiply $2$ by $- 2$ .

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

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

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

Step 1.3.4.2.1.3.7

Multiply $2$ by $- 1$ .

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

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

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

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

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

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

Step 1.3.4.2.1.4

Simplify by adding terms.

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Step 1.3.4.2.1.4.1

Add $- 1$ and $2$ .

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

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

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

Step 1.3.4.2.1.4.2

Subtract $2 C$ from $- 4 C$ .

$0 = \frac{1 - 6 C}{2}$

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

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

$0 = \frac{1 - 6 C}{2}$

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

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

$0 = \frac{1 - 6 C}{2}$

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

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

$0 = \frac{1 - 6 C}{2}$

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

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

$0 = \frac{1 - 6 C}{2}$

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

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

Step 1.3.5

Solve for $C$ in $0 = \frac{1 - 6 C}{2}$ .

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Step 1.3.5.1

Set the numerator equal to zero.

$1 - 6 C = 0$

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

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

Step 1.3.5.2

Solve the equation for $C$ .

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Step 1.3.5.2.1

Subtract $1$ from both sides of the equation.

$- 6 C = - 1$

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

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

Step 1.3.5.2.2

Divide each term in $- 6 C = - 1$ by $- 6$ and simplify.

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Step 1.3.5.2.2.1

Divide each term in $- 6 C = - 1$ by $- 6$ .

$\frac{- 6 C}{- 6} = \frac{- 1}{- 6}$

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

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

Step 1.3.5.2.2.2

Simplify the left side.

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Step 1.3.5.2.2.2.1

Cancel the common factor of $- 6$ .

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Step 1.3.5.2.2.2.1.1

Cancel the common factor.

$\frac{- 6 C}{- 6} = \frac{- 1}{- 6}$

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

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

Step 1.3.5.2.2.2.1.2

Divide $C$ by $1$ .

$C = \frac{- 1}{- 6}$

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

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

$C = \frac{- 1}{- 6}$

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

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

$C = \frac{- 1}{- 6}$

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

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

Step 1.3.5.2.2.3

Simplify the right side.

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Step 1.3.5.2.2.3.1

Dividing two negative values results in a positive value.

$C = \frac{1}{6}$

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

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

$C = \frac{1}{6}$

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

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

$C = \frac{1}{6}$

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

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

$C = \frac{1}{6}$

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

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

$C = \frac{1}{6}$

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

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

Step 1.3.6

Replace all occurrences of $C$ with $\frac{1}{6}$ in each equation.

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Step 1.3.6.1

Replace all occurrences of $C$ in $B = \frac{1}{2} - C$ with $\frac{1}{6}$ .

$B = \frac{1}{2} - (\frac{1}{6})$

$C = \frac{1}{6}$

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

Step 1.3.6.2

Simplify the right side.

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Step 1.3.6.2.1

Simplify $\frac{1}{2} - (\frac{1}{6})$ .

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Step 1.3.6.2.1.1

To write $\frac{1}{2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$B = \frac{1}{2} \cdot \frac{3}{3} - \frac{1}{6}$

$C = \frac{1}{6}$

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

Step 1.3.6.2.1.2

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.3.6.2.1.2.1

Multiply $\frac{1}{2}$ by $\frac{3}{3}$ .

$B = \frac{3}{2 \cdot 3} - \frac{1}{6}$

$C = \frac{1}{6}$

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

Step 1.3.6.2.1.2.2

Multiply $2$ by $3$ .

$B = \frac{3}{6} - \frac{1}{6}$

$C = \frac{1}{6}$

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

$B = \frac{3}{6} - \frac{1}{6}$

$C = \frac{1}{6}$

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

Step 1.3.6.2.1.3

Simplify the expression.

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Step 1.3.6.2.1.3.1

Combine the numerators over the common denominator.

$B = \frac{3 - 1}{6}$

$C = \frac{1}{6}$

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

Step 1.3.6.2.1.3.2

Subtract $1$ from $3$ .

$B = \frac{2}{6}$

$C = \frac{1}{6}$

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

$B = \frac{2}{6}$

$C = \frac{1}{6}$

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

Step 1.3.6.2.1.4

Cancel the common factor of $2$ and $6$ .

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Step 1.3.6.2.1.4.1

Factor $2$ out of $2$ .

$B = \frac{2 (1)}{6}$

$C = \frac{1}{6}$

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

Step 1.3.6.2.1.4.2

Cancel the common factors.

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Step 1.3.6.2.1.4.2.1

Factor $2$ out of $6$ .

$B = \frac{2 \cdot 1}{2 \cdot 3}$

$C = \frac{1}{6}$

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

Step 1.3.6.2.1.4.2.2

Cancel the common factor.

$B = \frac{2 \cdot 1}{2 \cdot 3}$

$C = \frac{1}{6}$

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

Step 1.3.6.2.1.4.2.3

Rewrite the expression.

$B = \frac{1}{3}$

$C = \frac{1}{6}$

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

$B = \frac{1}{3}$

$C = \frac{1}{6}$

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

$B = \frac{1}{3}$

$C = \frac{1}{6}$

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

$B = \frac{1}{3}$

$C = \frac{1}{6}$

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

$B = \frac{1}{3}$

$C = \frac{1}{6}$

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

$B = \frac{1}{3}$

$C = \frac{1}{6}$

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

Step 1.3.7

List all of the solutions.

$B = \frac{1}{3}, C = \frac{1}{6}, A = - \frac{1}{2}$

Step 1.4

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

$- \frac{\frac{1}{2}}{x} + \frac{\frac{1}{3}}{x - 1} + \frac{\frac{1}{6}}{x + 2}$

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

$- \frac{1}{2} \cdot \frac{1}{x} + \frac{\frac{1}{3}}{x - 1} + \frac{\frac{1}{6}}{x + 2}$

Step 1.5.2

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

$- \frac{1}{x \cdot 2} + \frac{\frac{1}{3}}{x - 1} + \frac{\frac{1}{6}}{x + 2}$

Step 1.5.3

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

$- \frac{1}{2 x} + \frac{\frac{1}{3}}{x - 1} + \frac{\frac{1}{6}}{x + 2}$

Step 1.5.4

Multiply the numerator by the reciprocal of the denominator.

$- \frac{1}{2 x} + \frac{1}{3} \cdot \frac{1}{x - 1} + \frac{\frac{1}{6}}{x + 2}$

Step 1.5.5

Multiply $\frac{1}{3}$ by $\frac{1}{x - 1}$ .

$- \frac{1}{2 x} + \frac{1}{3 (x - 1)} + \frac{\frac{1}{6}}{x + 2}$

Step 1.5.6

Multiply the numerator by the reciprocal of the denominator.

$- \frac{1}{2 x} + \frac{1}{3 (x - 1)} + \frac{1}{6} \cdot \frac{1}{x + 2}$

Step 1.5.7

Multiply $\frac{1}{6}$ by $\frac{1}{x + 2}$ .

$\int - \frac{1}{2 x} + \frac{1}{3 (x - 1)} + \frac{1}{6 (x + 2)} d x$

Step 2

Split the single integral into multiple integrals.

$\int - \frac{1}{2 x} d x + \int \frac{1}{3 (x - 1)} d x + \int \frac{1}{6 (x + 2)} d x$

Step 3

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

$- \int \frac{1}{2 x} d x + \int \frac{1}{3 (x - 1)} d x + \int \frac{1}{6 (x + 2)} d x$

Step 4

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

$- (\frac{1}{2} \int \frac{1}{x} d x) + \int \frac{1}{3 (x - 1)} d x + \int \frac{1}{6 (x + 2)} d x$

Step 5

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

$- \frac{1}{2} (\ln (| x |) + C) + \int \frac{1}{3 (x - 1)} d x + \int \frac{1}{6 (x + 2)} d x$

Step 6

Since $\frac{1}{3}$ is constant with respect to $x$ , move $\frac{1}{3}$ out of the integral.

$- \frac{1}{2} (\ln (| x |) + C) + \frac{1}{3} \int \frac{1}{x - 1} d x + \int \frac{1}{6 (x + 2)} d x$

Step 7

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

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Step 7.1

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

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Step 7.1.1

Differentiate $x - 1$ .

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

Step 7.1.2

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 7.1.3

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 7.1.4

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

$1 + 0$

Step 7.1.5

Add $1$ and $0$ .

$1$

Step 7.2

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

$- \frac{1}{2} (\ln (| x |) + C) + \frac{1}{3} \int \frac{1}{u_{1}} d u_{1} + \int \frac{1}{6 (x + 2)} d x$

Step 8

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

$- \frac{1}{2} (\ln (| x |) + C) + \frac{1}{3} (\ln (| u_{1} |) + C) + \int \frac{1}{6 (x + 2)} d x$

Step 9

Since $\frac{1}{6}$ is constant with respect to $x$ , move $\frac{1}{6}$ out of the integral.

$- \frac{1}{2} (\ln (| x |) + C) + \frac{1}{3} (\ln (| u_{1} |) + C) + \frac{1}{6} \int \frac{1}{x + 2} d x$

Step 10

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

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Step 10.1

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

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Step 10.1.1

Differentiate $x + 2$ .

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

Step 10.1.2

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 10.1.3

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 10.1.4

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

$1 + 0$

Step 10.1.5

Add $1$ and $0$ .

$1$

Step 10.2

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

$- \frac{1}{2} (\ln (| x |) + C) + \frac{1}{3} (\ln (| u_{1} |) + C) + \frac{1}{6} \int \frac{1}{u_{2}} d u_{2}$

Step 11

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

$- \frac{1}{2} (\ln (| x |) + C) + \frac{1}{3} (\ln (| u_{1} |) + C) + \frac{1}{6} (\ln (| u_{2} |) + C)$

Step 12

Simplify.

$- \frac{1}{2} \ln (| x |) + \frac{1}{3} \ln (| u_{1} |) + \frac{1}{6} \ln (| u_{2} |) + C$

Step 13

Substitute back in for each integration substitution variable.

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Step 13.1

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

$- \frac{1}{2} \ln (| x |) + \frac{1}{3} \ln (| x - 1 |) + \frac{1}{6} \ln (| u_{2} |) + C$

Step 13.2

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

$- \frac{1}{2} \ln (| x |) + \frac{1}{3} \ln (| x - 1 |) + \frac{1}{6} \ln (| x + 2 |) + C$