Calculus Examples

$\int \frac{12 + 10 x - 2 x^{2}}{x^{3} - 4 x} d x$

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 $2$ out of $12 + 10 x - 2 x^{2}$ .

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Step 1.1.1.1.1

Factor $2$ out of $12$ .

$\frac{2 (6) + 10 x - 2 x^{2}}{x^{3} - 4 x}$

Step 1.1.1.1.2

Factor $2$ out of $10 x$ .

$\frac{2 (6) + 2 (5 x) - 2 x^{2}}{x^{3} - 4 x}$

Step 1.1.1.1.3

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

$\frac{2 (6) + 2 (5 x) + 2 (- x^{2})}{x^{3} - 4 x}$

Step 1.1.1.1.4

Factor $2$ out of $2 (6) + 2 (5 x)$ .

$\frac{2 (6 + 5 x) + 2 (- x^{2})}{x^{3} - 4 x}$

Step 1.1.1.1.5

Factor $2$ out of $2 (6 + 5 x) + 2 (- x^{2})$ .

$\frac{2 (6 + 5 x - x^{2})}{x^{3} - 4 x}$

Step 1.1.1.2

Factor.

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Step 1.1.1.2.1

Factor by grouping.

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Step 1.1.1.2.1.1

Reorder terms.

$\frac{2 (- x^{2} + 5 x + 6)}{x^{3} - 4 x}$

Step 1.1.1.2.1.2

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 $a \cdot c = - 1 \cdot 6 = - 6$ and whose sum is $b = 5$ .

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Step 1.1.1.2.1.2.1

Factor $5$ out of $5 x$ .

$\frac{2 (- x^{2} + 5 (x) + 6)}{x^{3} - 4 x}$

Step 1.1.1.2.1.2.2

Rewrite $5$ as $- 1$ plus $6$

$\frac{2 (- x^{2} + (- 1 + 6) x + 6)}{x^{3} - 4 x}$

Step 1.1.1.2.1.2.3

Apply the distributive property.

$\frac{2 (- x^{2} - 1 x + 6 x + 6)}{x^{3} - 4 x}$

Step 1.1.1.2.1.3

Factor out the greatest common factor from each group.

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Step 1.1.1.2.1.3.1

Group the first two terms and the last two terms.

$\frac{2 ((- x^{2} - 1 x) + 6 x + 6)}{x^{3} - 4 x}$

Step 1.1.1.2.1.3.2

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

$\frac{2 (x (- x - 1) - 6 (- x - 1))}{x^{3} - 4 x}$

Step 1.1.1.2.1.4

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

$\frac{2 ((- x - 1) (x - 6))}{x^{3} - 4 x}$

Step 1.1.1.2.2

Remove unnecessary parentheses.

$\frac{2 (- x - 1) (x - 6)}{x^{3} - 4 x}$

Step 1.1.1.3

Factor $x$ out of $x^{3} - 4 x$ .

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Step 1.1.1.3.1

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

$\frac{2 (- x - 1) (x - 6)}{x \cdot x^{2} - 4 x}$

Step 1.1.1.3.2

Factor $x$ out of $- 4 x$ .

$\frac{2 (- x - 1) (x - 6)}{x \cdot x^{2} + x \cdot - 4}$

Step 1.1.1.3.3

Factor $x$ out of $x \cdot x^{2} + x \cdot - 4$ .

$\frac{2 (- x - 1) (x - 6)}{x (x^{2} - 4)}$

Step 1.1.1.4

Rewrite $4$ as $2^{2}$ .

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

Step 1.1.1.5

Factor.

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Step 1.1.1.5.1

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 $b = 2$ .

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

Step 1.1.1.5.2

Remove unnecessary parentheses.

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

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 $B$ .

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

Step 1.1.3

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

Step 1.1.4

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

$\frac{2 (- x - 1) (x - 6) (x (x + 2) (x - 2))}{x (x + 2) (x - 2)} = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.5

Simplify terms.

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Step 1.1.5.1

Cancel the common factor of $x$ .

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Step 1.1.5.1.1

Cancel the common factor.

$\frac{2 (- x - 1) (x - 6) (x (x + 2) (x - 2))}{x (x + 2) (x - 2)} = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.5.1.2

Rewrite the expression.

$\frac{2 (- x - 1) (x - 6) ((x + 2) (x - 2))}{(x + 2) (x - 2)} = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.5.2

Cancel the common factor of $x + 2$ .

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Step 1.1.5.2.1

Cancel the common factor.

$\frac{2 (- x - 1) (x - 6) ((x + 2) (x - 2))}{(x + 2) (x - 2)} = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.5.2.2

Rewrite the expression.

$\frac{2 (- x - 1) (x - 6) (x - 2)}{x - 2} = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.5.3

Cancel the common factor of $x - 2$ .

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Step 1.1.5.3.1

Cancel the common factor.

$\frac{2 (- x - 1) (x - 6) (x - 2)}{x - 2} = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.5.3.2

Divide $2 (- x - 1) (x - 6)$ by $1$ .

$2 (- x - 1) (x - 6) = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.5.4

Apply the distributive property.

$(2 (- x) + 2 \cdot - 1) (x - 6) = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.5.5

Multiply.

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Step 1.1.5.5.1

Multiply $- 1$ by $2$ .

$(- 2 x + 2 \cdot - 1) (x - 6) = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.5.5.2

Multiply $2$ by $- 1$ .

$(- 2 x - 2) (x - 6) = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.6

Expand $(- 2 x - 2) (x - 6)$ using the FOIL Method.

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Step 1.1.6.1

Apply the distributive property.

$- 2 x (x - 6) - 2 (x - 6) = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.6.2

Apply the distributive property.

$- 2 x \cdot x - 2 x \cdot - 6 - 2 (x - 6) = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.6.3

Apply the distributive property.

$- 2 x \cdot x - 2 x \cdot - 6 - 2 x - 2 \cdot - 6 = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.7

Simplify and combine like terms.

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Step 1.1.7.1

Simplify each term.

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Step 1.1.7.1.1

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

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Step 1.1.7.1.1.1

Move $x$ .

$- 2 (x \cdot x) - 2 x \cdot - 6 - 2 x - 2 \cdot - 6 = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.7.1.1.2

Multiply $x$ by $x$ .

$- 2 x^{2} - 2 x \cdot - 6 - 2 x - 2 \cdot - 6 = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.7.1.2

Multiply $- 6$ by $- 2$ .

$- 2 x^{2} + 12 x - 2 x - 2 \cdot - 6 = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.7.1.3

Multiply $- 2$ by $- 6$ .

$- 2 x^{2} + 12 x - 2 x + 12 = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.7.2

Subtract $2 x$ from $12 x$ .

$- 2 x^{2} + 10 x + 12 = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

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.

$- 2 x^{2} + 10 x + 12 = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.1.2

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

$- 2 x^{2} + 10 x + 12 = A ((x + 2) (x - 2)) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.2

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

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Step 1.1.8.2.1

Apply the distributive property.

$- 2 x^{2} + 10 x + 12 = A (x (x - 2) + 2 (x - 2)) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.2.2

Apply the distributive property.

$- 2 x^{2} + 10 x + 12 = A (x \cdot x + x \cdot - 2 + 2 (x - 2)) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.2.3

Apply the distributive property.

$- 2 x^{2} + 10 x + 12 = A (x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.3

Combine the opposite terms in $x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2$ .

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Step 1.1.8.3.1

Reorder the factors in the terms $x \cdot - 2$ and $2 x$ .

$- 2 x^{2} + 10 x + 12 = A (x \cdot x - 2 x + 2 x + 2 \cdot - 2) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.3.2

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

$- 2 x^{2} + 10 x + 12 = A (x \cdot x + 0 + 2 \cdot - 2) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.3.3

Add $x \cdot x$ and $0$ .

$- 2 x^{2} + 10 x + 12 = A (x \cdot x + 2 \cdot - 2) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.4

Simplify each term.

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Step 1.1.8.4.1

Multiply $x$ by $x$ .

$- 2 x^{2} + 10 x + 12 = A (x^{2} + 2 \cdot - 2) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.4.2

Multiply $2$ by $- 2$ .

$- 2 x^{2} + 10 x + 12 = A (x^{2} - 4) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.5

Apply the distributive property.

$- 2 x^{2} + 10 x + 12 = A x^{2} + A \cdot - 4 + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.6

Move $- 4$ to the left of $A$ .

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 \cdot A + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.7

Cancel the common factor of $x + 2$ .

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Step 1.1.8.7.1

Cancel the common factor.

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 A + \frac{B (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.7.2

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

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 A + (B) (x (x - 2)) + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.8

Apply the distributive property.

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 A + B (x \cdot x + x \cdot - 2) + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.9

Multiply $x$ by $x$ .

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 A + B (x^{2} + x \cdot - 2) + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.10

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

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 A + B (x^{2} - 2 \cdot x) + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.11

Apply the distributive property.

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 A + B x^{2} + B (- 2 x) + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.12

Rewrite using the commutative property of multiplication.

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.13

Cancel the common factor of $x - 2$ .

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Step 1.1.8.13.1

Cancel the common factor.

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + \frac{C (x (x + 2) (x - 2))}{x - 2}$

Step 1.1.8.13.2

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

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + (C) (x (x + 2))$

Step 1.1.8.14

Apply the distributive property.

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + C (x \cdot x + x \cdot 2)$

Step 1.1.8.15

Multiply $x$ by $x$ .

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + C (x^{2} + x \cdot 2)$

Step 1.1.8.16

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

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + C (x^{2} + 2 \cdot x)$

Step 1.1.8.17

Apply the distributive property.

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + C x^{2} + C (2 x)$

Step 1.1.8.18

Rewrite using the commutative property of multiplication.

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + C x^{2} + 2 C x$

Step 1.1.9

Simplify the expression.

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Step 1.1.9.1

Move $B$ .

$- 2 x^{2} + 10 x + 12 = A x^{2} - 4 A + B x^{2} - 2 x B + C x^{2} + 2 C x$

Step 1.1.9.2

Move $- 4 A$ .

$- 2 x^{2} + 10 x + 12 = A x^{2} + B x^{2} - 2 x B + C x^{2} + 2 C x - 4 A$

Step 1.1.9.3

Move $- 2 x B$ .

$- 2 x^{2} + 10 x + 12 = A x^{2} + B x^{2} + C x^{2} - 2 x B + 2 C x - 4 A$

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.

$- 2 = A + B + C$

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.

$10 = - 2 B + 2 C$

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.

$12 = - 4 A$

Step 1.2.4

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

$- 2 = A + B + C$

$10 = - 2 B + 2 C$

$12 = - 4 A$

$- 2 = A + B + C$

$10 = - 2 B + 2 C$

$12 = - 4 A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $A$ in $12 = - 4 A$ .

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Step 1.3.1.1

Rewrite the equation as $- 4 A = 12$ .

$- 4 A = 12$

$- 2 = A + B + C$

$10 = - 2 B + 2 C$

Step 1.3.1.2

Divide each term in $- 4 A = 12$ by $- 4$ and simplify.

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Step 1.3.1.2.1

Divide each term in $- 4 A = 12$ by $- 4$ .

$\frac{- 4 A}{- 4} = \frac{12}{- 4}$

$- 2 = A + B + C$

$10 = - 2 B + 2 C$

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 $- 4$ .

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Step 1.3.1.2.2.1.1

Cancel the common factor.

$\frac{- 4 A}{- 4} = \frac{12}{- 4}$

$- 2 = A + B + C$

$10 = - 2 B + 2 C$

Step 1.3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{12}{- 4}$

$- 2 = A + B + C$

$10 = - 2 B + 2 C$

$A = \frac{12}{- 4}$

$- 2 = A + B + C$

$10 = - 2 B + 2 C$

$A = \frac{12}{- 4}$

$- 2 = A + B + C$

$10 = - 2 B + 2 C$

Step 1.3.1.2.3

Simplify the right side.

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Step 1.3.1.2.3.1

Divide $12$ by $- 4$ .

$A = - 3$

$- 2 = A + B + C$

$10 = - 2 B + 2 C$

$A = - 3$

$- 2 = A + B + C$

$10 = - 2 B + 2 C$

$A = - 3$

$- 2 = A + B + C$

$10 = - 2 B + 2 C$

$A = - 3$

$- 2 = A + B + C$

$10 = - 2 B + 2 C$

Step 1.3.2

Replace all occurrences of $A$ with $- 3$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $A$ in $- 2 = A + B + C$ with $- 3$ .

$- 2 = (- 3) + B + C$

$A = - 3$

$10 = - 2 B + 2 C$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Remove parentheses.

$- 2 = - 3 + B + C$

$A = - 3$

$10 = - 2 B + 2 C$

$- 2 = - 3 + B + C$

$A = - 3$

$10 = - 2 B + 2 C$

$- 2 = - 3 + B + C$

$A = - 3$

$10 = - 2 B + 2 C$

Step 1.3.3

Solve for $B$ in $- 2 = - 3 + B + C$ .

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Step 1.3.3.1

Rewrite the equation as $- 3 + B + C = - 2$ .

$- 3 + B + C = - 2$

$A = - 3$

$10 = - 2 B + 2 C$

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 $3$ to both sides of the equation.

$B + C = - 2 + 3$

$A = - 3$

$10 = - 2 B + 2 C$

Step 1.3.3.2.2

Subtract $C$ from both sides of the equation.

$B = - 2 + 3 - C$

$A = - 3$

$10 = - 2 B + 2 C$

Step 1.3.3.2.3

Add $- 2$ and $3$ .

$B = 1 - C$

$A = - 3$

$10 = - 2 B + 2 C$

$B = 1 - C$

$A = - 3$

$10 = - 2 B + 2 C$

$B = 1 - C$

$A = - 3$

$10 = - 2 B + 2 C$

Step 1.3.4

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

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Step 1.3.4.1

Replace all occurrences of $B$ in $10 = - 2 B + 2 C$ with $1 - C$ .

$10 = - 2 (1 - C) + 2 C$

$B = 1 - C$

$A = - 3$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

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

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Step 1.3.4.2.1.1

Simplify each term.

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Step 1.3.4.2.1.1.1

Apply the distributive property.

$10 = - 2 \cdot 1 - 2 (- C) + 2 C$

$B = 1 - C$

$A = - 3$

Step 1.3.4.2.1.1.2

Multiply $- 2$ by $1$ .

$10 = - 2 - 2 (- C) + 2 C$

$B = 1 - C$

$A = - 3$

Step 1.3.4.2.1.1.3

Multiply $- 1$ by $- 2$ .

$10 = - 2 + 2 C + 2 C$

$B = 1 - C$

$A = - 3$

$10 = - 2 + 2 C + 2 C$

$B = 1 - C$

$A = - 3$

Step 1.3.4.2.1.2

Add $2 C$ and $2 C$ .

$10 = - 2 + 4 C$

$B = 1 - C$

$A = - 3$

$10 = - 2 + 4 C$

$B = 1 - C$

$A = - 3$

$10 = - 2 + 4 C$

$B = 1 - C$

$A = - 3$

$10 = - 2 + 4 C$

$B = 1 - C$

$A = - 3$

Step 1.3.5

Solve for $C$ in $10 = - 2 + 4 C$ .

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Step 1.3.5.1

Rewrite the equation as $- 2 + 4 C = 10$ .

$- 2 + 4 C = 10$

$B = 1 - C$

$A = - 3$

Step 1.3.5.2

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

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Step 1.3.5.2.1

Add $2$ to both sides of the equation.

$4 C = 10 + 2$

$B = 1 - C$

$A = - 3$

Step 1.3.5.2.2

Add $10$ and $2$ .

$4 C = 12$

$B = 1 - C$

$A = - 3$

$4 C = 12$

$B = 1 - C$

$A = - 3$

Step 1.3.5.3

Divide each term in $4 C = 12$ by $4$ and simplify.

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Step 1.3.5.3.1

Divide each term in $4 C = 12$ by $4$ .

$\frac{4 C}{4} = \frac{12}{4}$

$B = 1 - C$

$A = - 3$

Step 1.3.5.3.2

Simplify the left side.

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Step 1.3.5.3.2.1

Cancel the common factor of $4$ .

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Step 1.3.5.3.2.1.1

Cancel the common factor.

$\frac{4 C}{4} = \frac{12}{4}$

$B = 1 - C$

$A = - 3$

Step 1.3.5.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{12}{4}$

$B = 1 - C$

$A = - 3$

$C = \frac{12}{4}$

$B = 1 - C$

$A = - 3$

$C = \frac{12}{4}$

$B = 1 - C$

$A = - 3$

Step 1.3.5.3.3

Simplify the right side.

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Step 1.3.5.3.3.1

Divide $12$ by $4$ .

$C = 3$

$B = 1 - C$

$A = - 3$

$C = 3$

$B = 1 - C$

$A = - 3$

$C = 3$

$B = 1 - C$

$A = - 3$

$C = 3$

$B = 1 - C$

$A = - 3$

Step 1.3.6

Replace all occurrences of $C$ with $3$ in each equation.

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Step 1.3.6.1

Replace all occurrences of $C$ in $B = 1 - C$ with $3$ .

$B = 1 - (3)$

$C = 3$

$A = - 3$

Step 1.3.6.2

Simplify the right side.

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Step 1.3.6.2.1

Simplify $1 - (3)$ .

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Step 1.3.6.2.1.1

Multiply $- 1$ by $3$ .

$B = 1 - 3$

$C = 3$

$A = - 3$

Step 1.3.6.2.1.2

Subtract $3$ from $1$ .

$B = - 2$

$C = 3$

$A = - 3$

$B = - 2$

$C = 3$

$A = - 3$

$B = - 2$

$C = 3$

$A = - 3$

$B = - 2$

$C = 3$

$A = - 3$

Step 1.3.7

List all of the solutions.

$B = - 2, C = 3, A = - 3$

Step 1.4

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

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

Step 1.5

Simplify.

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Step 1.5.1

Move the negative in front of the fraction.

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

Step 1.5.2

Move the negative in front of the fraction.

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

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

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

Step 4

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

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

Step 5

Multiply $3$ by $- 1$ .

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

Step 6

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

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

Step 7

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

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

Step 8

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

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

Step 9

Multiply $2$ by $- 1$ .

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

Step 10

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

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Step 10.1

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

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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 $u_{1}$ and $d u_{1}$ .

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

Step 11

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

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

Step 12

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

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

Step 13

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

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Step 13.1

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

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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 $d u_{2}$ .

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

Step 14

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

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

Step 15

Simplify.

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

Step 16

Substitute back in for each integration substitution variable.

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Step 16.1

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

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

Step 16.2

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

$- 3 \ln (| x |) - 2 \ln (| x + 2 |) + 3 \ln (| x - 2 |) + C$