Calculus Examples

$\int \frac{2 x^{2} - 9 x - 35}{(x + 1) (x + 2) (x + 3)} d x$

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 by grouping.

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Step 1.1.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 35 = - 70$ and whose sum is $b = - 9$ .

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Step 1.1.1.1.1

Factor $- 9$ out of $- 9 x$ .

$\frac{2 x^{2} - 9 (x) - 35}{(x + 1) (x + 2) (x + 3)}$

Step 1.1.1.1.2

Rewrite $- 9$ as $5$ plus $- 14$

$\frac{2 x^{2} + (5 - 14) x - 35}{(x + 1) (x + 2) (x + 3)}$

Step 1.1.1.1.3

Apply the distributive property.

$\frac{2 x^{2} + 5 x - 14 x - 35}{(x + 1) (x + 2) (x + 3)}$

Step 1.1.1.2

Factor out the greatest common factor from each group.

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Step 1.1.1.2.1

Group the first two terms and the last two terms.

$\frac{(2 x^{2} + 5 x) - 14 x - 35}{(x + 1) (x + 2) (x + 3)}$

Step 1.1.1.2.2

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

$\frac{x (2 x + 5) - 7 (2 x + 5)}{(x + 1) (x + 2) (x + 3)}$

Step 1.1.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x + 5$ .

$\frac{(2 x + 5) (x - 7)}{(x + 1) (x + 2) (x + 3)}$

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 $A$ .

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

Step 1.1.3

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

Step 1.1.4

$\frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}$

Step 1.1.5

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

$\frac{(2 x + 5) (x - 7) (x + 1) (x + 2) (x + 3)}{(x + 1) (x + 2) (x + 3)} = \frac{(A) (x + 1) (x + 2) (x + 3)}{x + 1} + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.6

Reduce the expression by cancelling the common factors.

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Step 1.1.6.1

Cancel the common factor of $x + 1$ .

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Step 1.1.6.1.1

Cancel the common factor.

Step 1.1.6.1.2

Rewrite the expression.

$\frac{((2 x + 5) (x - 7) (x + 2)) (x + 3)}{(x + 2) (x + 3)} = \frac{(A) (x + 1) (x + 2) (x + 3)}{x + 1} + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.6.2

Cancel the common factor of $x + 2$ .

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Step 1.1.6.2.1

Cancel the common factor.

$\frac{(2 x + 5) (x - 7) (x + 2) (x + 3)}{(x + 2) (x + 3)} = \frac{(A) (x + 1) (x + 2) (x + 3)}{x + 1} + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.6.2.2

Rewrite the expression.

$\frac{((2 x + 5) (x - 7)) (x + 3)}{x + 3} = \frac{(A) (x + 1) (x + 2) (x + 3)}{x + 1} + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.6.3

Cancel the common factor of $x + 3$ .

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Step 1.1.6.3.1

Cancel the common factor.

$\frac{(2 x + 5) (x - 7) (x + 3)}{x + 3} = \frac{(A) (x + 1) (x + 2) (x + 3)}{x + 1} + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.6.3.2

Divide $(2 x + 5) (x - 7)$ by $1$ .

$(2 x + 5) (x - 7) = \frac{(A) (x + 1) (x + 2) (x + 3)}{x + 1} + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.7

Expand $(2 x + 5) (x - 7)$ using the FOIL Method.

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Step 1.1.7.1

Apply the distributive property.

$2 x (x - 7) + 5 (x - 7) = \frac{(A) (x + 1) (x + 2) (x + 3)}{x + 1} + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.7.2

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 7 + 5 (x - 7) = \frac{(A) (x + 1) (x + 2) (x + 3)}{x + 1} + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.7.3

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 7 + 5 x + 5 \cdot - 7 = \frac{(A) (x + 1) (x + 2) (x + 3)}{x + 1} + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.8

Simplify and combine like terms.

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Step 1.1.8.1

Simplify each term.

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Step 1.1.8.1.1

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

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Step 1.1.8.1.1.1

Move $x$ .

$2 (x \cdot x) + 2 x \cdot - 7 + 5 x + 5 \cdot - 7 = \frac{(A) (x + 1) (x + 2) (x + 3)}{x + 1} + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.8.1.1.2

Multiply $x$ by $x$ .

$2 x^{2} + 2 x \cdot - 7 + 5 x + 5 \cdot - 7 = \frac{(A) (x + 1) (x + 2) (x + 3)}{x + 1} + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.8.1.2

Multiply $- 7$ by $2$ .

$2 x^{2} - 14 x + 5 x + 5 \cdot - 7 = \frac{(A) (x + 1) (x + 2) (x + 3)}{x + 1} + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.8.1.3

Multiply $5$ by $- 7$ .

$2 x^{2} - 14 x + 5 x - 35 = \frac{(A) (x + 1) (x + 2) (x + 3)}{x + 1} + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.8.2

Add $- 14 x$ and $5 x$ .

$2 x^{2} - 9 x - 35 = \frac{(A) (x + 1) (x + 2) (x + 3)}{x + 1} + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9

Simplify each term.

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Step 1.1.9.1

Cancel the common factor of $x + 1$ .

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Step 1.1.9.1.1

Cancel the common factor.

$2 x^{2} - 9 x - 35 = \frac{A (x + 1) (x + 2) (x + 3)}{x + 1} + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.1.2

Divide $((A) (x + 2)) (x + 3)$ by $1$ .

$2 x^{2} - 9 x - 35 = ((A) (x + 2)) (x + 3) + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.2

Apply the distributive property.

$2 x^{2} - 9 x - 35 = (A x + A \cdot 2) (x + 3) + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.3

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

$2 x^{2} - 9 x - 35 = (A x + 2 \cdot A) (x + 3) + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.4

Expand $(A x + 2 A) (x + 3)$ using the FOIL Method.

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Step 1.1.9.4.1

Apply the distributive property.

$2 x^{2} - 9 x - 35 = A x (x + 3) + 2 A (x + 3) + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.4.2

Apply the distributive property.

$2 x^{2} - 9 x - 35 = A x \cdot x + A x \cdot 3 + 2 A (x + 3) + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.4.3

Apply the distributive property.

$2 x^{2} - 9 x - 35 = A x \cdot x + A x \cdot 3 + 2 A x + 2 A \cdot 3 + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.5

Simplify and combine like terms.

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Step 1.1.9.5.1

Simplify each term.

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Step 1.1.9.5.1.1

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

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Step 1.1.9.5.1.1.1

Move $x$ .

$2 x^{2} - 9 x - 35 = A (x \cdot x) + A x \cdot 3 + 2 A x + 2 A \cdot 3 + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.5.1.1.2

Multiply $x$ by $x$ .

$2 x^{2} - 9 x - 35 = A x^{2} + A x \cdot 3 + 2 A x + 2 A \cdot 3 + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.5.1.2

Move $3$ to the left of $A x$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 3 \cdot (A x) + 2 A x + 2 A \cdot 3 + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.5.1.3

Multiply $3$ by $2$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 3 A x + 2 A x + 6 A + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.5.2

Add $3 A x$ and $2 A x$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.6

Cancel the common factor of $x + 2$ .

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Step 1.1.9.6.1

Cancel the common factor.

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + \frac{(B) (x + 1) (x + 2) (x + 3)}{x + 2} + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.6.2

Divide $(B) (x + 1) (x + 3)$ by $1$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + (B) (x + 1) (x + 3) + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.7

Apply the distributive property.

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + (B x + B \cdot 1) (x + 3) + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.8

Multiply $B$ by $1$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + (B x + B) (x + 3) + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.9

Expand $(B x + B) (x + 3)$ using the FOIL Method.

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Step 1.1.9.9.1

Apply the distributive property.

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x (x + 3) + B (x + 3) + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.9.2

Apply the distributive property.

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x \cdot x + B x \cdot 3 + B (x + 3) + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.9.3

Apply the distributive property.

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x \cdot x + B x \cdot 3 + B x + B \cdot 3 + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.10

Simplify and combine like terms.

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Step 1.1.9.10.1

Simplify each term.

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Step 1.1.9.10.1.1

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

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Step 1.1.9.10.1.1.1

Move $x$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B (x \cdot x) + B x \cdot 3 + B x + B \cdot 3 + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.10.1.1.2

Multiply $x$ by $x$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + B x \cdot 3 + B x + B \cdot 3 + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.10.1.2

Move $3$ to the left of $B x$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + 3 \cdot (B x) + B x + B \cdot 3 + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.10.1.3

Move $3$ to the left of $B$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + 3 B x + B x + 3 B + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.10.2

Add $3 B x$ and $B x$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + 4 B x + 3 B + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.11

Cancel the common factor of $x + 3$ .

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Step 1.1.9.11.1

Cancel the common factor.

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + 4 B x + 3 B + \frac{(C) (x + 1) (x + 2) (x + 3)}{x + 3}$

Step 1.1.9.11.2

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

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + 4 B x + 3 B + (C) (x + 1) (x + 2)$

Step 1.1.9.12

Apply the distributive property.

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + 4 B x + 3 B + (C x + C \cdot 1) (x + 2)$

Step 1.1.9.13

Multiply $C$ by $1$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + 4 B x + 3 B + (C x + C) (x + 2)$

Step 1.1.9.14

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

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Step 1.1.9.14.1

Apply the distributive property.

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + 4 B x + 3 B + C x (x + 2) + C (x + 2)$

Step 1.1.9.14.2

Apply the distributive property.

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + 4 B x + 3 B + C x \cdot x + C x \cdot 2 + C (x + 2)$

Step 1.1.9.14.3

Apply the distributive property.

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + 4 B x + 3 B + C x \cdot x + C x \cdot 2 + C x + C \cdot 2$

Step 1.1.9.15

Simplify and combine like terms.

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Step 1.1.9.15.1

Simplify each term.

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Step 1.1.9.15.1.1

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

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Step 1.1.9.15.1.1.1

Move $x$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + 4 B x + 3 B + C (x \cdot x) + C x \cdot 2 + C x + C \cdot 2$

Step 1.1.9.15.1.1.2

Multiply $x$ by $x$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + 4 B x + 3 B + C x^{2} + C x \cdot 2 + C x + C \cdot 2$

Step 1.1.9.15.1.2

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

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + 4 B x + 3 B + C x^{2} + 2 \cdot (C x) + C x + C \cdot 2$

Step 1.1.9.15.1.3

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

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + 4 B x + 3 B + C x^{2} + 2 C x + C x + 2 C$

Step 1.1.9.15.2

Add $2 C x$ and $C x$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 A x + 6 A + B x^{2} + 4 B x + 3 B + C x^{2} + 3 C x + 2 C$

Step 1.1.10

Simplify the expression.

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Step 1.1.10.1

Move $A$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 x A + 6 A + B x^{2} + 4 B x + 3 B + C x^{2} + 3 C x + 2 C$

Step 1.1.10.2

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

$2 x^{2} - 9 x - 35 = A x^{2} + 5 x A + 6 A + B x^{2} + 4 B x + 3 B + C x^{2} + 3 C x + 2 C$

Step 1.1.10.3

Move $B$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 x A + 6 A + B x^{2} + 4 x B + 3 B + C x^{2} + 3 C x + 2 C$

Step 1.1.10.4

Move $3 B$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 x A + 6 A + B x^{2} + 4 x B + C x^{2} + 3 C x + 3 B + 2 C$

Step 1.1.10.5

Move $6 A$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 x A + B x^{2} + 4 x B + C x^{2} + 3 C x + 6 A + 3 B + 2 C$

Step 1.1.10.6

Move $4 x B$ .

$2 x^{2} - 9 x - 35 = A x^{2} + 5 x A + B x^{2} + C x^{2} + 4 x B + 3 C x + 6 A + 3 B + 2 C$

Step 1.1.10.7

Move $5 x A$ .

$2 x^{2} - 9 x - 35 = A x^{2} + B x^{2} + C x^{2} + 5 x A + 4 x B + 3 C x + 6 A + 3 B + 2 C$

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.

$- 9 = 5 A + 4 B + 3 C$

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.

$- 35 = 6 A + 3 B + 2 C$

Step 1.2.4

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

$2 = A + B + C$

$- 9 = 5 A + 4 B + 3 C$

$- 35 = 6 A + 3 B + 2 C$

$2 = A + B + C$

$- 9 = 5 A + 4 B + 3 C$

$- 35 = 6 A + 3 B + 2 C$

Step 1.3

Solve the system of equations.

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Step 1.3.1

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

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Step 1.3.1.1

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

$A + B + C = 2$

$- 9 = 5 A + 4 B + 3 C$

$- 35 = 6 A + 3 B + 2 C$

Step 1.3.1.2

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

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Step 1.3.1.2.1

Subtract $B$ from both sides of the equation.

$A + C = 2 - B$

$- 9 = 5 A + 4 B + 3 C$

$- 35 = 6 A + 3 B + 2 C$

Step 1.3.1.2.2

Subtract $C$ from both sides of the equation.

$A = 2 - B - C$

$- 9 = 5 A + 4 B + 3 C$

$- 35 = 6 A + 3 B + 2 C$

$A = 2 - B - C$

$- 9 = 5 A + 4 B + 3 C$

$- 35 = 6 A + 3 B + 2 C$

$A = 2 - B - C$

$- 9 = 5 A + 4 B + 3 C$

$- 35 = 6 A + 3 B + 2 C$

Step 1.3.2

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

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Step 1.3.2.1

Replace all occurrences of $A$ in $- 9 = 5 A + 4 B + 3 C$ with $2 - B - C$ .

$- 9 = 5 (2 - B - C) + 4 B + 3 C$

$A = 2 - B - C$

$- 35 = 6 A + 3 B + 2 C$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Simplify $5 (2 - B - C) + 4 B + 3 C$ .

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Step 1.3.2.2.1.1

Simplify each term.

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Step 1.3.2.2.1.1.1

Apply the distributive property.

$- 9 = 5 \cdot 2 + 5 (- B) + 5 (- C) + 4 B + 3 C$

$A = 2 - B - C$

$- 35 = 6 A + 3 B + 2 C$

Step 1.3.2.2.1.1.2

Simplify.

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Step 1.3.2.2.1.1.2.1

Multiply $5$ by $2$ .

$- 9 = 10 + 5 (- B) + 5 (- C) + 4 B + 3 C$

$A = 2 - B - C$

$- 35 = 6 A + 3 B + 2 C$

Step 1.3.2.2.1.1.2.2

Multiply $- 1$ by $5$ .

$- 9 = 10 - 5 B + 5 (- C) + 4 B + 3 C$

$A = 2 - B - C$

$- 35 = 6 A + 3 B + 2 C$

Step 1.3.2.2.1.1.2.3

Multiply $- 1$ by $5$ .

$- 9 = 10 - 5 B - 5 C + 4 B + 3 C$

$A = 2 - B - C$

$- 35 = 6 A + 3 B + 2 C$

$- 9 = 10 - 5 B - 5 C + 4 B + 3 C$

$A = 2 - B - C$

$- 35 = 6 A + 3 B + 2 C$

$- 9 = 10 - 5 B - 5 C + 4 B + 3 C$

$A = 2 - B - C$

$- 35 = 6 A + 3 B + 2 C$

Step 1.3.2.2.1.2

Simplify by adding terms.

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Step 1.3.2.2.1.2.1

Add $- 5 B$ and $4 B$ .

$- 9 = 10 - B - 5 C + 3 C$

$A = 2 - B - C$

$- 35 = 6 A + 3 B + 2 C$

Step 1.3.2.2.1.2.2

Add $- 5 C$ and $3 C$ .

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$- 35 = 6 A + 3 B + 2 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$- 35 = 6 A + 3 B + 2 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$- 35 = 6 A + 3 B + 2 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$- 35 = 6 A + 3 B + 2 C$

Step 1.3.2.3

Replace all occurrences of $A$ in $- 35 = 6 A + 3 B + 2 C$ with $2 - B - C$ .

$- 35 = 6 (2 - B - C) + 3 B + 2 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

Step 1.3.2.4

Simplify the right side.

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Step 1.3.2.4.1

Simplify $6 (2 - B - C) + 3 B + 2 C$ .

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Step 1.3.2.4.1.1

Simplify each term.

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Step 1.3.2.4.1.1.1

Apply the distributive property.

$- 35 = 6 \cdot 2 + 6 (- B) + 6 (- C) + 3 B + 2 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

Step 1.3.2.4.1.1.2

Simplify.

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Step 1.3.2.4.1.1.2.1

Multiply $6$ by $2$ .

$- 35 = 12 + 6 (- B) + 6 (- C) + 3 B + 2 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

Step 1.3.2.4.1.1.2.2

Multiply $- 1$ by $6$ .

$- 35 = 12 - 6 B + 6 (- C) + 3 B + 2 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

Step 1.3.2.4.1.1.2.3

Multiply $- 1$ by $6$ .

$- 35 = 12 - 6 B - 6 C + 3 B + 2 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$- 35 = 12 - 6 B - 6 C + 3 B + 2 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$- 35 = 12 - 6 B - 6 C + 3 B + 2 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

Step 1.3.2.4.1.2

Simplify by adding terms.

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Step 1.3.2.4.1.2.1

Add $- 6 B$ and $3 B$ .

$- 35 = 12 - 3 B - 6 C + 2 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

Step 1.3.2.4.1.2.2

Add $- 6 C$ and $2 C$ .

$- 35 = 12 - 3 B - 4 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$- 35 = 12 - 3 B - 4 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$- 35 = 12 - 3 B - 4 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$- 35 = 12 - 3 B - 4 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$- 35 = 12 - 3 B - 4 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

Step 1.3.3

Solve for $B$ in $- 35 = 12 - 3 B - 4 C$ .

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Step 1.3.3.1

Rewrite the equation as $12 - 3 B - 4 C = - 35$ .

$12 - 3 B - 4 C = - 35$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

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

Subtract $12$ from both sides of the equation.

$- 3 B - 4 C = - 35 - 12$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

Step 1.3.3.2.2

Add $4 C$ to both sides of the equation.

$- 3 B = - 35 - 12 + 4 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

Step 1.3.3.2.3

Subtract $12$ from $- 35$ .

$- 3 B = - 47 + 4 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$- 3 B = - 47 + 4 C$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

Step 1.3.3.3

Divide each term in $- 3 B = - 47 + 4 C$ by $- 3$ and simplify.

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Step 1.3.3.3.1

Divide each term in $- 3 B = - 47 + 4 C$ by $- 3$ .

$\frac{- 3 B}{- 3} = \frac{- 47}{- 3} + \frac{4 C}{- 3}$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

Step 1.3.3.3.2

Simplify the left side.

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Step 1.3.3.3.2.1

Cancel the common factor of $- 3$ .

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Step 1.3.3.3.2.1.1

Cancel the common factor.

$\frac{- 3 B}{- 3} = \frac{- 47}{- 3} + \frac{4 C}{- 3}$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

Step 1.3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{- 47}{- 3} + \frac{4 C}{- 3}$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$B = \frac{- 47}{- 3} + \frac{4 C}{- 3}$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$B = \frac{- 47}{- 3} + \frac{4 C}{- 3}$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Simplify each term.

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Step 1.3.3.3.3.1.1

Dividing two negative values results in a positive value.

$B = \frac{47}{3} + \frac{4 C}{- 3}$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

Step 1.3.3.3.3.1.2

Move the negative in front of the fraction.

$B = \frac{47}{3} - \frac{4 C}{3}$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$- 9 = 10 - B - 2 C$

$A = 2 - B - C$

Step 1.3.4

Replace all occurrences of $B$ with $\frac{47}{3} - \frac{4 C}{3}$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $B$ in $- 9 = 10 - B - 2 C$ with $\frac{47}{3} - \frac{4 C}{3}$ .

$- 9 = 10 - (\frac{47}{3} - \frac{4 C}{3}) - 2 C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $10 - (\frac{47}{3} - \frac{4 C}{3}) - 2 C$ .

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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.

$- 9 = 10 - \frac{47}{3} + \frac{4 C}{3} - 2 C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.1.2

Multiply $- - \frac{4 C}{3}$ .

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Step 1.3.4.2.1.1.2.1

Multiply $- 1$ by $- 1$ .

$- 9 = 10 - \frac{47}{3} + 1 (\frac{4 C}{3}) - 2 C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.1.2.2

Multiply $\frac{4 C}{3}$ by $1$ .

$- 9 = 10 - \frac{47}{3} + \frac{4 C}{3} - 2 C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

$- 9 = 10 - \frac{47}{3} + \frac{4 C}{3} - 2 C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

$- 9 = 10 - \frac{47}{3} + \frac{4 C}{3} - 2 C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.2

To write $10$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- 9 = 10 \cdot \frac{3}{3} - \frac{47}{3} + \frac{4 C}{3} - 2 C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.3

Combine $10$ and $\frac{3}{3}$ .

$- 9 = \frac{10 \cdot 3}{3} - \frac{47}{3} + \frac{4 C}{3} - 2 C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.4

Combine the numerators over the common denominator.

$- 9 = \frac{10 \cdot 3 - 47}{3} + \frac{4 C}{3} - 2 C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.5

Simplify the numerator.

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Step 1.3.4.2.1.5.1

Multiply $10$ by $3$ .

$- 9 = \frac{30 - 47}{3} + \frac{4 C}{3} - 2 C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.5.2

Subtract $47$ from $30$ .

$- 9 = \frac{- 17}{3} + \frac{4 C}{3} - 2 C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

$- 9 = \frac{- 17}{3} + \frac{4 C}{3} - 2 C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.6

Move the negative in front of the fraction.

$- 9 = - \frac{17}{3} + \frac{4 C}{3} - 2 C$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.7

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

$- 9 = - \frac{17}{3} + \frac{4 C}{3} - 2 C \cdot \frac{3}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.8

Combine $- 2 C$ and $\frac{3}{3}$ .

$- 9 = - \frac{17}{3} + \frac{4 C}{3} + \frac{- 2 C \cdot 3}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.9

Combine the numerators over the common denominator.

$- 9 = - \frac{17}{3} + \frac{4 C - 2 C \cdot 3}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.10

Combine the numerators over the common denominator.

$- 9 = \frac{- 17 + 4 C - 2 C \cdot 3}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.11

Multiply $3$ by $- 2$ .

$- 9 = \frac{- 17 + 4 C - 6 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.12

Subtract $6 C$ from $4 C$ .

$- 9 = \frac{- 17 - 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.13

Rewrite $- 17$ as $- 1 (17)$ .

$- 9 = \frac{- 1 \cdot 17 - 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.14

Factor $- 1$ out of $- 2 C$ .

$- 9 = \frac{- 1 \cdot 17 - (2 C)}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.15

Factor $- 1$ out of $- 1 (17) - (2 C)$ .

$- 9 = \frac{- 1 (17 + 2 C)}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.2.1.16

Move the negative in front of the fraction.

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - B - C$

Step 1.3.4.3

Replace all occurrences of $B$ in $A = 2 - B - C$ with $\frac{47}{3} - \frac{4 C}{3}$ .

$A = 2 - (\frac{47}{3} - \frac{4 C}{3}) - C$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4

Simplify the right side.

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Step 1.3.4.4.1

Simplify $2 - (\frac{47}{3} - \frac{4 C}{3}) - C$ .

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Step 1.3.4.4.1.1

Simplify each term.

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Step 1.3.4.4.1.1.1

Apply the distributive property.

$A = 2 - \frac{47}{3} + \frac{4 C}{3} - C$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.1.2

Multiply $- - \frac{4 C}{3}$ .

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Step 1.3.4.4.1.1.2.1

Multiply $- 1$ by $- 1$ .

$A = 2 - \frac{47}{3} + 1 (\frac{4 C}{3}) - C$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.1.2.2

Multiply $\frac{4 C}{3}$ by $1$ .

$A = 2 - \frac{47}{3} + \frac{4 C}{3} - C$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - \frac{47}{3} + \frac{4 C}{3} - C$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = 2 - \frac{47}{3} + \frac{4 C}{3} - C$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.2

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

$A = 2 \cdot \frac{3}{3} - \frac{47}{3} + \frac{4 C}{3} - C$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.3

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

$A = \frac{2 \cdot 3}{3} - \frac{47}{3} + \frac{4 C}{3} - C$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.4

Combine the numerators over the common denominator.

$A = \frac{2 \cdot 3 - 47}{3} + \frac{4 C}{3} - C$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.5

Simplify the numerator.

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Step 1.3.4.4.1.5.1

Multiply $2$ by $3$ .

$A = \frac{6 - 47}{3} + \frac{4 C}{3} - C$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.5.2

Subtract $47$ from $6$ .

$A = \frac{- 41}{3} + \frac{4 C}{3} - C$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = \frac{- 41}{3} + \frac{4 C}{3} - C$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.6

Move the negative in front of the fraction.

$A = - \frac{41}{3} + \frac{4 C}{3} - C$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.7

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

$A = - \frac{41}{3} + \frac{4 C}{3} - C \cdot \frac{3}{3}$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.8

Combine $- C$ and $\frac{3}{3}$ .

$A = - \frac{41}{3} + \frac{4 C}{3} + \frac{- C \cdot 3}{3}$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.9

Combine the numerators over the common denominator.

$A = - \frac{41}{3} + \frac{4 C - C \cdot 3}{3}$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.10

Combine the numerators over the common denominator.

$A = \frac{- 41 + 4 C - C \cdot 3}{3}$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.11

Multiply $3$ by $- 1$ .

$A = \frac{- 41 + 4 C - 3 C}{3}$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.12

Subtract $3 C$ from $4 C$ .

$A = \frac{- 41 + C}{3}$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.13

Rewrite $- 41$ as $- 1 (41)$ .

$A = \frac{- 1 \cdot 41 + C}{3}$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.14

Factor $- 1$ out of $C$ .

$A = \frac{- 1 \cdot 41 - 1 (- C)}{3}$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.15

Factor $- 1$ out of $- 1 (41) - 1 (- C)$ .

$A = \frac{- 1 (41 - C)}{3}$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.4.4.1.16

Move the negative in front of the fraction.

$A = - \frac{41 - C}{3}$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = - \frac{41 - C}{3}$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = - \frac{41 - C}{3}$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = - \frac{41 - C}{3}$

$- 9 = - \frac{17 + 2 C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.5

Solve for $C$ in $- 9 = - \frac{17 + 2 C}{3}$ .

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Step 1.3.5.1

Rewrite the equation as $- \frac{17 + 2 C}{3} = - 9$ .

$- \frac{17 + 2 C}{3} = - 9$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.5.2

Multiply both sides of the equation by $- 3$ .

$- 3 (- \frac{17 + 2 C}{3}) = - 3 \cdot - 9$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.5.3

Simplify both sides of the equation.

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Step 1.3.5.3.1

Simplify the left side.

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Step 1.3.5.3.1.1

Simplify $- 3 (- \frac{17 + 2 C}{3})$ .

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Step 1.3.5.3.1.1.1

Cancel the common factor of $3$ .

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Step 1.3.5.3.1.1.1.1

Move the leading negative in $- \frac{17 + 2 C}{3}$ into the numerator.

$- 3 \frac{- (17 + 2 C)}{3} = - 3 \cdot - 9$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.5.3.1.1.1.2

Factor $3$ out of $- 3$ .

$3 (- 1) (\frac{- (17 + 2 C)}{3}) = - 3 \cdot - 9$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.5.3.1.1.1.3

Cancel the common factor.

$3 \cdot (- 1 \frac{- (17 + 2 C)}{3}) = - 3 \cdot - 9$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.5.3.1.1.1.4

Rewrite the expression.

$17 + 2 C = - 3 \cdot - 9$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$17 + 2 C = - 3 \cdot - 9$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.5.3.1.1.2

Multiply.

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Step 1.3.5.3.1.1.2.1

Multiply $- 1$ by $- 1$ .

$1 (17 + 2 C) = - 3 \cdot - 9$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.5.3.1.1.2.2

Multiply $17 + 2 C$ by $1$ .

$17 + 2 C = - 3 \cdot - 9$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$17 + 2 C = - 3 \cdot - 9$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$17 + 2 C = - 3 \cdot - 9$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$17 + 2 C = - 3 \cdot - 9$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.5.3.2

Simplify the right side.

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Step 1.3.5.3.2.1

Multiply $- 3$ by $- 9$ .

$17 + 2 C = 27$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$17 + 2 C = 27$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$17 + 2 C = 27$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.5.4

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

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Step 1.3.5.4.1

Subtract $17$ from both sides of the equation.

$2 C = 27 - 17$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.5.4.2

Subtract $17$ from $27$ .

$2 C = 10$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$2 C = 10$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.5.5

Divide each term in $2 C = 10$ by $2$ and simplify.

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Step 1.3.5.5.1

Divide each term in $2 C = 10$ by $2$ .

$\frac{2 C}{2} = \frac{10}{2}$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.5.5.2

Simplify the left side.

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Step 1.3.5.5.2.1

Cancel the common factor of $2$ .

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Step 1.3.5.5.2.1.1

Cancel the common factor.

$\frac{2 C}{2} = \frac{10}{2}$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.5.5.2.1.2

Divide $C$ by $1$ .

$C = \frac{10}{2}$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$C = \frac{10}{2}$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$C = \frac{10}{2}$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.5.5.3

Simplify the right side.

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Step 1.3.5.5.3.1

Divide $10$ by $2$ .

$C = 5$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$C = 5$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$C = 5$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

$C = 5$

$A = - \frac{41 - C}{3}$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.6

Replace all occurrences of $C$ with $5$ in each equation.

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Step 1.3.6.1

Replace all occurrences of $C$ in $A = - \frac{41 - C}{3}$ with $5$ .

$A = - \frac{41 - (5)}{3}$

$C = 5$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.6.2

Simplify the right side.

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Step 1.3.6.2.1

Simplify $- \frac{41 - (5)}{3}$ .

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Step 1.3.6.2.1.1

Simplify the numerator.

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Step 1.3.6.2.1.1.1

Multiply $- 1$ by $5$ .

$A = - \frac{41 - 5}{3}$

$C = 5$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.6.2.1.1.2

Subtract $5$ from $41$ .

$A = - \frac{36}{3}$

$C = 5$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = - \frac{36}{3}$

$C = 5$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.6.2.1.2

Simplify the expression.

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Step 1.3.6.2.1.2.1

Divide $36$ by $3$ .

$A = - 1 \cdot 12$

$C = 5$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.6.2.1.2.2

Multiply $- 1$ by $12$ .

$A = - 12$

$C = 5$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = - 12$

$C = 5$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = - 12$

$C = 5$

$B = \frac{47}{3} - \frac{4 C}{3}$

$A = - 12$

$C = 5$

$B = \frac{47}{3} - \frac{4 C}{3}$

Step 1.3.6.3

Replace all occurrences of $C$ in $B = \frac{47}{3} - \frac{4 C}{3}$ with $5$ .

$B = \frac{47}{3} - \frac{4 (5)}{3}$

$A = - 12$

$C = 5$

Step 1.3.6.4

Simplify the right side.

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Step 1.3.6.4.1

Simplify $\frac{47}{3} - \frac{4 (5)}{3}$ .

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Step 1.3.6.4.1.1

Combine the numerators over the common denominator.

$B = \frac{47 - 4 \cdot 5}{3}$

$A = - 12$

$C = 5$

Step 1.3.6.4.1.2

Simplify the expression.

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Step 1.3.6.4.1.2.1

Multiply $- 4$ by $5$ .

$B = \frac{47 - 20}{3}$

$A = - 12$

$C = 5$

Step 1.3.6.4.1.2.2

Subtract $20$ from $47$ .

$B = \frac{27}{3}$

$A = - 12$

$C = 5$

Step 1.3.6.4.1.2.3

Divide $27$ by $3$ .

$B = 9$

$A = - 12$

$C = 5$

$B = 9$

$A = - 12$

$C = 5$

$B = 9$

$A = - 12$

$C = 5$

$B = 9$

$A = - 12$

$C = 5$

$B = 9$

$A = - 12$

$C = 5$

Step 1.3.7

List all of the solutions.

$B = 9, A = - 12, C = 5$

Step 1.4

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

$\frac{- 12}{x + 1} + \frac{9}{x + 2} + \frac{5}{x + 3}$

Step 1.5

Move the negative in front of the fraction.

$\int - \frac{12}{x + 1} + \frac{9}{x + 2} + \frac{5}{x + 3} d x$

Step 2

Split the single integral into multiple integrals.

$\int - \frac{12}{x + 1} d x + \int \frac{9}{x + 2} d x + \int \frac{5}{x + 3} d x$

Step 3

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

$- \int \frac{12}{x + 1} d x + \int \frac{9}{x + 2} d x + \int \frac{5}{x + 3} d x$

Step 4

Since $12$ is constant with respect to $x$ , move $12$ out of the integral.

$- (12 \int \frac{1}{x + 1} d x) + \int \frac{9}{x + 2} d x + \int \frac{5}{x + 3} d x$

Step 5

Multiply $12$ by $- 1$ .

$- 12 \int \frac{1}{x + 1} d x + \int \frac{9}{x + 2} d x + \int \frac{5}{x + 3} d x$

Step 6

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

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Step 6.1

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

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Step 6.1.1

Differentiate $x + 1$ .

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

Step 6.1.2

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 6.1.3

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 6.1.4

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

$1 + 0$

Step 6.1.5

Add $1$ and $0$ .

$1$

Step 6.2

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

$- 12 \int \frac{1}{u_{1}} d u_{1} + \int \frac{9}{x + 2} d x + \int \frac{5}{x + 3} d x$

Step 7

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

$- 12 (\ln (| u_{1} |) + C) + \int \frac{9}{x + 2} d x + \int \frac{5}{x + 3} d x$

Step 8

Since $9$ is constant with respect to $x$ , move $9$ out of the integral.

$- 12 (\ln (| u_{1} |) + C) + 9 \int \frac{1}{x + 2} d x + \int \frac{5}{x + 3} d x$

Step 9

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

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Step 9.1

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

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Step 9.1.1

Differentiate $x + 2$ .

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

Step 9.1.2

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 9.1.3

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

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

Step 9.1.4

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

$1 + 0$

Step 9.1.5

Add $1$ and $0$ .

$1$

Step 9.2

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

$- 12 (\ln (| u_{1} |) + C) + 9 \int \frac{1}{u_{2}} d u_{2} + \int \frac{5}{x + 3} d x$

Step 10

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

$- 12 (\ln (| u_{1} |) + C) + 9 (\ln (| u_{2} |) + C) + \int \frac{5}{x + 3} d x$

Step 11

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

$- 12 (\ln (| u_{1} |) + C) + 9 (\ln (| u_{2} |) + C) + 5 \int \frac{1}{x + 3} d x$

Step 12

Let $u_{3} = x + 3$ . Then $d u_{3} = d x$ . Rewrite using $u_{3}$ and $d$ $u_{3}$ .

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Step 12.1

Let $u_{3} = x + 3$ . Find $\frac{d u_{3}}{d x}$ .

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Step 12.1.1

Differentiate $x + 3$ .

$\frac{d}{d x} [x + 3]$

Step 12.1.2

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [3]$

Step 12.1.3

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

$1 + \frac{d}{d x} [3]$

Step 12.1.4

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$1 + 0$

Step 12.1.5

Add $1$ and $0$ .

$1$

Step 12.2

Rewrite the problem using $u_{3}$ and $d u_{3}$ .

$- 12 (\ln (| u_{1} |) + C) + 9 (\ln (| u_{2} |) + C) + 5 \int \frac{1}{u_{3}} d u_{3}$

Step 13

The integral of $\frac{1}{u_{3}}$ with respect to $u_{3}$ is $\ln (| u_{3} |)$ .

$- 12 (\ln (| u_{1} |) + C) + 9 (\ln (| u_{2} |) + C) + 5 (\ln (| u_{3} |) + C)$

Step 14

Simplify.

$- 12 \ln (| u_{1} |) + 9 \ln (| u_{2} |) + 5 \ln (| u_{3} |) + C$

Step 15

Substitute back in for each integration substitution variable.

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Step 15.1

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

$- 12 \ln (| x + 1 |) + 9 \ln (| u_{2} |) + 5 \ln (| u_{3} |) + C$

Step 15.2

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

$- 12 \ln (| x + 1 |) + 9 \ln (| x + 2 |) + 5 \ln (| u_{3} |) + C$

Step 15.3

Replace all occurrences of $u_{3}$ with $x + 3$ .

$- 12 \ln (| x + 1 |) + 9 \ln (| x + 2 |) + 5 \ln (| x + 3 |) + C$