Calculus Examples

$\int \frac{4 x^{2} + 38 x + 9}{{(x + 3)}^{2} (x + 5)} d x$

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

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 $A$ .

$\frac{A}{{(x + 3)}^{2}}$

Step 1.1.2

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

Step 1.1.3

$\frac{A}{{(x + 3)}^{2}} + \frac{B}{x + 3} + \frac{C}{x + 5}$

Step 1.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is ${(x + 3)}^{2} (x + 5)$ .

$\frac{(4 x^{2} + 38 x + 9) {(x + 3)}^{2} (x + 5)}{{(x + 3)}^{2} (x + 5)} = \frac{(A) {(x + 3)}^{2} (x + 5)}{{(x + 3)}^{2}} + \frac{(B) {(x + 3)}^{2} (x + 5)}{x + 3} + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.5

Cancel the common factor of ${(x + 3)}^{2}$ .

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Step 1.1.5.1

Cancel the common factor.

Step 1.1.5.2

Rewrite the expression.

$\frac{(4 x^{2} + 38 x + 9) (x + 5)}{x + 5} = \frac{(A) {(x + 3)}^{2} (x + 5)}{{(x + 3)}^{2}} + \frac{(B) {(x + 3)}^{2} (x + 5)}{x + 3} + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.6

Cancel the common factor of $x + 5$ .

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Step 1.1.6.1

Cancel the common factor.

$\frac{(4 x^{2} + 38 x + 9) (x + 5)}{x + 5} = \frac{(A) {(x + 3)}^{2} (x + 5)}{{(x + 3)}^{2}} + \frac{(B) {(x + 3)}^{2} (x + 5)}{x + 3} + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.6.2

Divide $4 x^{2} + 38 x + 9$ by $1$ .

$4 x^{2} + 38 x + 9 = \frac{(A) {(x + 3)}^{2} (x + 5)}{{(x + 3)}^{2}} + \frac{(B) {(x + 3)}^{2} (x + 5)}{x + 3} + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

Cancel the common factor of ${(x + 3)}^{2}$ .

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Step 1.1.7.1.1

Cancel the common factor.

$4 x^{2} + 38 x + 9 = \frac{A {(x + 3)}^{2} (x + 5)}{{(x + 3)}^{2}} + \frac{(B) {(x + 3)}^{2} (x + 5)}{x + 3} + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.1.2

Divide $(A) (x + 5)$ by $1$ .

$4 x^{2} + 38 x + 9 = (A) (x + 5) + \frac{(B) {(x + 3)}^{2} (x + 5)}{x + 3} + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.2

Apply the distributive property.

$4 x^{2} + 38 x + 9 = A x + A \cdot 5 + \frac{(B) {(x + 3)}^{2} (x + 5)}{x + 3} + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.3

Move $5$ to the left of $A$ .

$4 x^{2} + 38 x + 9 = A x + 5 \cdot A + \frac{(B) {(x + 3)}^{2} (x + 5)}{x + 3} + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.4

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

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Step 1.1.7.4.1

Factor $x + 3$ out of $(B) {(x + 3)}^{2} (x + 5)$ .

$4 x^{2} + 38 x + 9 = A x + 5 A + \frac{(x + 3) (B (x + 3) (x + 5))}{x + 3} + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.4.2

Cancel the common factors.

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Step 1.1.7.4.2.1

Multiply by $1$ .

$4 x^{2} + 38 x + 9 = A x + 5 A + \frac{(x + 3) (B (x + 3) (x + 5))}{(x + 3) \cdot 1} + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.4.2.2

Cancel the common factor.

$4 x^{2} + 38 x + 9 = A x + 5 A + \frac{(x + 3) (B (x + 3) (x + 5))}{(x + 3) \cdot 1} + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.4.2.3

Rewrite the expression.

$4 x^{2} + 38 x + 9 = A x + 5 A + \frac{B (x + 3) (x + 5)}{1} + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.4.2.4

Divide $B (x + 3) (x + 5)$ by $1$ .

$4 x^{2} + 38 x + 9 = A x + 5 A + B (x + 3) (x + 5) + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.5

Apply the distributive property.

$4 x^{2} + 38 x + 9 = A x + 5 A + (B x + B \cdot 3) (x + 5) + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.6

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

$4 x^{2} + 38 x + 9 = A x + 5 A + (B x + 3 \cdot B) (x + 5) + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.7

Expand $(B x + 3 B) (x + 5)$ using the FOIL Method.

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Step 1.1.7.7.1

Apply the distributive property.

$4 x^{2} + 38 x + 9 = A x + 5 A + B x (x + 5) + 3 B (x + 5) + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.7.2

Apply the distributive property.

$4 x^{2} + 38 x + 9 = A x + 5 A + B x \cdot x + B x \cdot 5 + 3 B (x + 5) + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.7.3

Apply the distributive property.

$4 x^{2} + 38 x + 9 = A x + 5 A + B x \cdot x + B x \cdot 5 + 3 B x + 3 B \cdot 5 + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.8

Simplify and combine like terms.

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Step 1.1.7.8.1

Simplify each term.

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Step 1.1.7.8.1.1

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

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Step 1.1.7.8.1.1.1

Move $x$ .

$4 x^{2} + 38 x + 9 = A x + 5 A + B (x \cdot x) + B x \cdot 5 + 3 B x + 3 B \cdot 5 + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.8.1.1.2

Multiply $x$ by $x$ .

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + B x \cdot 5 + 3 B x + 3 B \cdot 5 + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.8.1.2

Move $5$ to the left of $B x$ .

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 5 \cdot (B x) + 3 B x + 3 B \cdot 5 + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.8.1.3

Multiply $5$ by $3$ .

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 5 B x + 3 B x + 15 B + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.8.2

Add $5 B x$ and $3 B x$ .

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 B x + 15 B + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.9

Cancel the common factor of $x + 5$ .

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Step 1.1.7.9.1

Cancel the common factor.

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 B x + 15 B + \frac{(C) {(x + 3)}^{2} (x + 5)}{x + 5}$

Step 1.1.7.9.2

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

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 B x + 15 B + (C) {(x + 3)}^{2}$

Step 1.1.7.10

Rewrite ${(x + 3)}^{2}$ as $(x + 3) (x + 3)$ .

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 B x + 15 B + C ((x + 3) (x + 3))$

Step 1.1.7.11

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

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Step 1.1.7.11.1

Apply the distributive property.

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 B x + 15 B + C (x (x + 3) + 3 (x + 3))$

Step 1.1.7.11.2

Apply the distributive property.

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 B x + 15 B + C (x \cdot x + x \cdot 3 + 3 (x + 3))$

Step 1.1.7.11.3

Apply the distributive property.

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 B x + 15 B + C (x \cdot x + x \cdot 3 + 3 x + 3 \cdot 3)$

Step 1.1.7.12

Simplify and combine like terms.

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Step 1.1.7.12.1

Simplify each term.

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Step 1.1.7.12.1.1

Multiply $x$ by $x$ .

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 B x + 15 B + C (x^{2} + x \cdot 3 + 3 x + 3 \cdot 3)$

Step 1.1.7.12.1.2

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

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 B x + 15 B + C (x^{2} + 3 \cdot x + 3 x + 3 \cdot 3)$

Step 1.1.7.12.1.3

Multiply $3$ by $3$ .

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 B x + 15 B + C (x^{2} + 3 x + 3 x + 9)$

Step 1.1.7.12.2

Add $3 x$ and $3 x$ .

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 B x + 15 B + C (x^{2} + 6 x + 9)$

Step 1.1.7.13

Apply the distributive property.

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 B x + 15 B + C x^{2} + C (6 x) + C \cdot 9$

Step 1.1.7.14

Simplify.

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Step 1.1.7.14.1

Rewrite using the commutative property of multiplication.

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 B x + 15 B + C x^{2} + 6 C x + C \cdot 9$

Step 1.1.7.14.2

Move $9$ to the left of $C$ .

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 B x + 15 B + C x^{2} + 6 C x + 9 \cdot C$

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 B x + 15 B + C x^{2} + 6 C x + 9 C$

Step 1.1.8

Simplify the expression.

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Step 1.1.8.1

Move $B$ .

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 x B + 15 B + C x^{2} + 6 C x + 9 C$

Step 1.1.8.2

Move $15 B$ .

$4 x^{2} + 38 x + 9 = A x + 5 A + B x^{2} + 8 x B + C x^{2} + 6 C x + 15 B + 9 C$

Step 1.1.8.3

Move $5 A$ .

$4 x^{2} + 38 x + 9 = A x + B x^{2} + 8 x B + C x^{2} + 6 C x + 5 A + 15 B + 9 C$

Step 1.1.8.4

Move $8 x B$ .

$4 x^{2} + 38 x + 9 = A x + B x^{2} + C x^{2} + 8 x B + 6 C x + 5 A + 15 B + 9 C$

Step 1.1.8.5

Move $A x$ .

$4 x^{2} + 38 x + 9 = B x^{2} + C x^{2} + A x + 8 x B + 6 C x + 5 A + 15 B + 9 C$

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.

$4 = B + C$

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.

$38 = A + 8 B + 6 C$

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.

$9 = 5 A + 15 B + 9 C$

Step 1.2.4

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

$4 = B + C$

$38 = A + 8 B + 6 C$

$9 = 5 A + 15 B + 9 C$

$4 = B + C$

$38 = A + 8 B + 6 C$

$9 = 5 A + 15 B + 9 C$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $B$ in $4 = B + C$ .

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Step 1.3.1.1

Rewrite the equation as $B + C = 4$ .

$B + C = 4$

$38 = A + 8 B + 6 C$

$9 = 5 A + 15 B + 9 C$

Step 1.3.1.2

Subtract $C$ from both sides of the equation.

$B = 4 - C$

$38 = A + 8 B + 6 C$

$9 = 5 A + 15 B + 9 C$

$B = 4 - C$

$38 = A + 8 B + 6 C$

$9 = 5 A + 15 B + 9 C$

Step 1.3.2

Replace all occurrences of $B$ with $4 - C$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $B$ in $38 = A + 8 B + 6 C$ with $4 - C$ .

$38 = A + 8 (4 - C) + 6 C$

$B = 4 - C$

$9 = 5 A + 15 B + 9 C$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Simplify $A + 8 (4 - C) + 6 C$ .

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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.

$38 = A + 8 \cdot 4 + 8 (- C) + 6 C$

$B = 4 - C$

$9 = 5 A + 15 B + 9 C$

Step 1.3.2.2.1.1.2

Multiply $8$ by $4$ .

$38 = A + 32 + 8 (- C) + 6 C$

$B = 4 - C$

$9 = 5 A + 15 B + 9 C$

Step 1.3.2.2.1.1.3

Multiply $- 1$ by $8$ .

$38 = A + 32 - 8 C + 6 C$

$B = 4 - C$

$9 = 5 A + 15 B + 9 C$

$38 = A + 32 - 8 C + 6 C$

$B = 4 - C$

$9 = 5 A + 15 B + 9 C$

Step 1.3.2.2.1.2

Add $- 8 C$ and $6 C$ .

$38 = A + 32 - 2 C$

$B = 4 - C$

$9 = 5 A + 15 B + 9 C$

$38 = A + 32 - 2 C$

$B = 4 - C$

$9 = 5 A + 15 B + 9 C$

$38 = A + 32 - 2 C$

$B = 4 - C$

$9 = 5 A + 15 B + 9 C$

Step 1.3.2.3

Replace all occurrences of $B$ in $9 = 5 A + 15 B + 9 C$ with $4 - C$ .

$9 = 5 A + 15 (4 - C) + 9 C$

$38 = A + 32 - 2 C$

$B = 4 - C$

Step 1.3.2.4

Simplify the right side.

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Step 1.3.2.4.1

Simplify $5 A + 15 (4 - C) + 9 C$ .

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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.

$9 = 5 A + 15 \cdot 4 + 15 (- C) + 9 C$

$38 = A + 32 - 2 C$

$B = 4 - C$

Step 1.3.2.4.1.1.2

Multiply $15$ by $4$ .

$9 = 5 A + 60 + 15 (- C) + 9 C$

$38 = A + 32 - 2 C$

$B = 4 - C$

Step 1.3.2.4.1.1.3

Multiply $- 1$ by $15$ .

$9 = 5 A + 60 - 15 C + 9 C$

$38 = A + 32 - 2 C$

$B = 4 - C$

$9 = 5 A + 60 - 15 C + 9 C$

$38 = A + 32 - 2 C$

$B = 4 - C$

Step 1.3.2.4.1.2

Add $- 15 C$ and $9 C$ .

$9 = 5 A + 60 - 6 C$

$38 = A + 32 - 2 C$

$B = 4 - C$

$9 = 5 A + 60 - 6 C$

$38 = A + 32 - 2 C$

$B = 4 - C$

$9 = 5 A + 60 - 6 C$

$38 = A + 32 - 2 C$

$B = 4 - C$

$9 = 5 A + 60 - 6 C$

$38 = A + 32 - 2 C$

$B = 4 - C$

Step 1.3.3

Reorder $4$ and $- C$ .

$B = - C + 4$

$9 = 5 A + 60 - 6 C$

$38 = A + 32 - 2 C$

Step 1.3.4

Solve for $A$ in $38 = A + 32 - 2 C$ .

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Step 1.3.4.1

Rewrite the equation as $A + 32 - 2 C = 38$ .

$A + 32 - 2 C = 38$

$B = - C + 4$

$9 = 5 A + 60 - 6 C$

Step 1.3.4.2

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

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Step 1.3.4.2.1

Subtract $32$ from both sides of the equation.

$A - 2 C = 38 - 32$

$B = - C + 4$

$9 = 5 A + 60 - 6 C$

Step 1.3.4.2.2

Add $2 C$ to both sides of the equation.

$A = 38 - 32 + 2 C$

$B = - C + 4$

$9 = 5 A + 60 - 6 C$

Step 1.3.4.2.3

Subtract $32$ from $38$ .

$A = 6 + 2 C$

$B = - C + 4$

$9 = 5 A + 60 - 6 C$

$A = 6 + 2 C$

$B = - C + 4$

$9 = 5 A + 60 - 6 C$

$A = 6 + 2 C$

$B = - C + 4$

$9 = 5 A + 60 - 6 C$

Step 1.3.5

Replace all occurrences of $A$ with $6 + 2 C$ in each equation.

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Step 1.3.5.1

Replace all occurrences of $A$ in $9 = 5 A + 60 - 6 C$ with $6 + 2 C$ .

$9 = 5 (6 + 2 C) + 60 - 6 C$

$A = 6 + 2 C$

$B = - C + 4$

Step 1.3.5.2

Simplify the right side.

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Step 1.3.5.2.1

Simplify $5 (6 + 2 C) + 60 - 6 C$ .

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Step 1.3.5.2.1.1

Simplify each term.

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Step 1.3.5.2.1.1.1

Apply the distributive property.

$9 = 5 \cdot 6 + 5 (2 C) + 60 - 6 C$

$A = 6 + 2 C$

$B = - C + 4$

Step 1.3.5.2.1.1.2

Multiply $5$ by $6$ .

$9 = 30 + 5 (2 C) + 60 - 6 C$

$A = 6 + 2 C$

$B = - C + 4$

Step 1.3.5.2.1.1.3

Multiply $2$ by $5$ .

$9 = 30 + 10 C + 60 - 6 C$

$A = 6 + 2 C$

$B = - C + 4$

$9 = 30 + 10 C + 60 - 6 C$

$A = 6 + 2 C$

$B = - C + 4$

Step 1.3.5.2.1.2

Simplify by adding terms.

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Step 1.3.5.2.1.2.1

Add $30$ and $60$ .

$9 = 10 C + 90 - 6 C$

$A = 6 + 2 C$

$B = - C + 4$

Step 1.3.5.2.1.2.2

Subtract $6 C$ from $10 C$ .

$9 = 4 C + 90$

$A = 6 + 2 C$

$B = - C + 4$

$9 = 4 C + 90$

$A = 6 + 2 C$

$B = - C + 4$

$9 = 4 C + 90$

$A = 6 + 2 C$

$B = - C + 4$

$9 = 4 C + 90$

$A = 6 + 2 C$

$B = - C + 4$

$9 = 4 C + 90$

$A = 6 + 2 C$

$B = - C + 4$

Step 1.3.6

Solve for $C$ in $9 = 4 C + 90$ .

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Step 1.3.6.1

Rewrite the equation as $4 C + 90 = 9$ .

$4 C + 90 = 9$

$A = 6 + 2 C$

$B = - C + 4$

Step 1.3.6.2

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

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Step 1.3.6.2.1

Subtract $90$ from both sides of the equation.

$4 C = 9 - 90$

$A = 6 + 2 C$

$B = - C + 4$

Step 1.3.6.2.2

Subtract $90$ from $9$ .

$4 C = - 81$

$A = 6 + 2 C$

$B = - C + 4$

$4 C = - 81$

$A = 6 + 2 C$

$B = - C + 4$

Step 1.3.6.3

Divide each term in $4 C = - 81$ by $4$ and simplify.

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Step 1.3.6.3.1

Divide each term in $4 C = - 81$ by $4$ .

$\frac{4 C}{4} = \frac{- 81}{4}$

$A = 6 + 2 C$

$B = - C + 4$

Step 1.3.6.3.2

Simplify the left side.

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Step 1.3.6.3.2.1

Cancel the common factor of $4$ .

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Step 1.3.6.3.2.1.1

Cancel the common factor.

$\frac{4 C}{4} = \frac{- 81}{4}$

$A = 6 + 2 C$

$B = - C + 4$

Step 1.3.6.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{- 81}{4}$

$A = 6 + 2 C$

$B = - C + 4$

$C = \frac{- 81}{4}$

$A = 6 + 2 C$

$B = - C + 4$

$C = \frac{- 81}{4}$

$A = 6 + 2 C$

$B = - C + 4$

Step 1.3.6.3.3

Simplify the right side.

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Step 1.3.6.3.3.1

Move the negative in front of the fraction.

$C = - \frac{81}{4}$

$A = 6 + 2 C$

$B = - C + 4$

$C = - \frac{81}{4}$

$A = 6 + 2 C$

$B = - C + 4$

$C = - \frac{81}{4}$

$A = 6 + 2 C$

$B = - C + 4$

$C = - \frac{81}{4}$

$A = 6 + 2 C$

$B = - C + 4$

Step 1.3.7

Replace all occurrences of $C$ with $- \frac{81}{4}$ in each equation.

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Step 1.3.7.1

Replace all occurrences of $C$ in $A = 6 + 2 C$ with $- \frac{81}{4}$ .

$A = 6 + 2 (- \frac{81}{4})$

$C = - \frac{81}{4}$

$B = - C + 4$

Step 1.3.7.2

Simplify the right side.

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Step 1.3.7.2.1

Simplify $6 + 2 (- \frac{81}{4})$ .

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Step 1.3.7.2.1.1

Simplify each term.

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Step 1.3.7.2.1.1.1

Cancel the common factor of $2$ .

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Step 1.3.7.2.1.1.1.1

Move the leading negative in $- \frac{81}{4}$ into the numerator.

$A = 6 + 2 (\frac{- 81}{4})$

$C = - \frac{81}{4}$

$B = - C + 4$

Step 1.3.7.2.1.1.1.2

Factor $2$ out of $4$ .

$A = 6 + 2 (\frac{- 81}{2 (2)})$

$C = - \frac{81}{4}$

$B = - C + 4$

Step 1.3.7.2.1.1.1.3

Cancel the common factor.

$A = 6 + 2 (\frac{- 81}{2 \cdot 2})$

$C = - \frac{81}{4}$

$B = - C + 4$

Step 1.3.7.2.1.1.1.4

Rewrite the expression.

$A = 6 + \frac{- 81}{2}$

$C = - \frac{81}{4}$

$B = - C + 4$

$A = 6 + \frac{- 81}{2}$

$C = - \frac{81}{4}$

$B = - C + 4$

Step 1.3.7.2.1.1.2

Move the negative in front of the fraction.

$A = 6 - \frac{81}{2}$

$C = - \frac{81}{4}$

$B = - C + 4$

$A = 6 - \frac{81}{2}$

$C = - \frac{81}{4}$

$B = - C + 4$

Step 1.3.7.2.1.2

To write $6$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$A = 6 \cdot \frac{2}{2} - \frac{81}{2}$

$C = - \frac{81}{4}$

$B = - C + 4$

Step 1.3.7.2.1.3

Combine $6$ and $\frac{2}{2}$ .

$A = \frac{6 \cdot 2}{2} - \frac{81}{2}$

$C = - \frac{81}{4}$

$B = - C + 4$

Step 1.3.7.2.1.4

Combine the numerators over the common denominator.

$A = \frac{6 \cdot 2 - 81}{2}$

$C = - \frac{81}{4}$

$B = - C + 4$

Step 1.3.7.2.1.5

Simplify the numerator.

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Step 1.3.7.2.1.5.1

Multiply $6$ by $2$ .

$A = \frac{12 - 81}{2}$

$C = - \frac{81}{4}$

$B = - C + 4$

Step 1.3.7.2.1.5.2

Subtract $81$ from $12$ .

$A = \frac{- 69}{2}$

$C = - \frac{81}{4}$

$B = - C + 4$

$A = \frac{- 69}{2}$

$C = - \frac{81}{4}$

$B = - C + 4$

Step 1.3.7.2.1.6

Move the negative in front of the fraction.

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

$B = - C + 4$

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

$B = - C + 4$

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

$B = - C + 4$

Step 1.3.7.3

Replace all occurrences of $C$ in $B = - C + 4$ with $- \frac{81}{4}$ .

$B = - (- \frac{81}{4}) + 4$

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

Step 1.3.7.4

Simplify the right side.

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Step 1.3.7.4.1

Simplify $- (- \frac{81}{4}) + 4$ .

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Step 1.3.7.4.1.1

Multiply $- (- \frac{81}{4})$ .

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Step 1.3.7.4.1.1.1

Multiply $- 1$ by $- 1$ .

$B = 1 (\frac{81}{4}) + 4$

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

Step 1.3.7.4.1.1.2

Multiply $\frac{81}{4}$ by $1$ .

$B = \frac{81}{4} + 4$

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

$B = \frac{81}{4} + 4$

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

Step 1.3.7.4.1.2

To write $4$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$B = \frac{81}{4} + 4 \cdot \frac{4}{4}$

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

Step 1.3.7.4.1.3

Combine $4$ and $\frac{4}{4}$ .

$B = \frac{81}{4} + \frac{4 \cdot 4}{4}$

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

Step 1.3.7.4.1.4

Combine the numerators over the common denominator.

$B = \frac{81 + 4 \cdot 4}{4}$

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

Step 1.3.7.4.1.5

Simplify the numerator.

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Step 1.3.7.4.1.5.1

Multiply $4$ by $4$ .

$B = \frac{81 + 16}{4}$

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

Step 1.3.7.4.1.5.2

Add $81$ and $16$ .

$B = \frac{97}{4}$

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

$B = \frac{97}{4}$

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

$B = \frac{97}{4}$

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

$B = \frac{97}{4}$

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

$B = \frac{97}{4}$

$A = - \frac{69}{2}$

$C = - \frac{81}{4}$

Step 1.3.8

List all of the solutions.

$B = \frac{97}{4}, A = - \frac{69}{2}, C = - \frac{81}{4}$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{{(x + 3)}^{2}} + \frac{B}{x + 3} + \frac{C}{x + 5}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{- \frac{69}{2}}{{(x + 3)}^{2}} + \frac{\frac{97}{4}}{x + 3} + \frac{- \frac{81}{4}}{x + 5}$

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

$- \frac{69}{2} \cdot \frac{1}{{(x + 3)}^{2}} + \frac{\frac{97}{4}}{x + 3} + \frac{- \frac{81}{4}}{x + 5}$

Step 1.5.2

Multiply $\frac{1}{{(x + 3)}^{2}}$ by $\frac{69}{2}$ .

$- \frac{69}{{(x + 3)}^{2} \cdot 2} + \frac{\frac{97}{4}}{x + 3} + \frac{- \frac{81}{4}}{x + 5}$

Step 1.5.3

Move $2$ to the left of ${(x + 3)}^{2}$ .

$- \frac{69}{2 {(x + 3)}^{2}} + \frac{\frac{97}{4}}{x + 3} + \frac{- \frac{81}{4}}{x + 5}$

Step 1.5.4

Multiply the numerator by the reciprocal of the denominator.

$- \frac{69}{2 {(x + 3)}^{2}} + \frac{97}{4} \cdot \frac{1}{x + 3} + \frac{- \frac{81}{4}}{x + 5}$

Step 1.5.5

Multiply $\frac{97}{4}$ by $\frac{1}{x + 3}$ .

$- \frac{69}{2 {(x + 3)}^{2}} + \frac{97}{4 (x + 3)} + \frac{- \frac{81}{4}}{x + 5}$

Step 1.5.6

Multiply the numerator by the reciprocal of the denominator.

$- \frac{69}{2 {(x + 3)}^{2}} + \frac{97}{4 (x + 3)} - \frac{81}{4} \cdot \frac{1}{x + 5}$

Step 1.5.7

Multiply $\frac{1}{x + 5}$ by $\frac{81}{4}$ .

$- \frac{69}{2 {(x + 3)}^{2}} + \frac{97}{4 (x + 3)} - \frac{81}{(x + 5) \cdot 4}$

Step 1.5.8

Move $4$ to the left of $x + 5$ .

$\int - \frac{69}{2 {(x + 3)}^{2}} + \frac{97}{4 (x + 3)} - \frac{81}{4 (x + 5)} d x$

Step 2

Split the single integral into multiple integrals.

$\int - \frac{69}{2 {(x + 3)}^{2}} d x + \int \frac{97}{4 (x + 3)} d x + \int - \frac{81}{4 (x + 5)} d x$

Step 3

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

$- \int \frac{69}{2 {(x + 3)}^{2}} d x + \int \frac{97}{4 (x + 3)} d x + \int - \frac{81}{4 (x + 5)} d x$

Step 4

Since $\frac{69}{2}$ is constant with respect to $x$ , move $\frac{69}{2}$ out of the integral.

$- (\frac{69}{2} \int \frac{1}{{(x + 3)}^{2}} d x) + \int \frac{97}{4 (x + 3)} d x + \int - \frac{81}{4 (x + 5)} d x$

Step 5

Let $u_{1} = x + 3$ . Then $d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 5.1

Let $u_{1} = x + 3$ . Find $\frac{d u_{1}}{d x}$ .

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Step 5.1.1

Differentiate $x + 3$ .

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

Step 5.1.2

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 5.1.3

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 5.1.4

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

$1 + 0$

Step 5.1.5

Add $1$ and $0$ .

$1$

Step 5.2

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

$- \frac{69}{2} \int \frac{1}{{u_{1}}^{2}} d u_{1} + \int \frac{97}{4 (x + 3)} d x + \int - \frac{81}{4 (x + 5)} d x$

Step 6

Apply basic rules of exponents.

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Step 6.1

Move ${u_{1}}^{2}$ out of the denominator by raising it to the $- 1$ power.

$- \frac{69}{2} \int {({u_{1}}^{2})}^{- 1} d u_{1} + \int \frac{97}{4 (x + 3)} d x + \int - \frac{81}{4 (x + 5)} d x$

Step 6.2

Multiply the exponents in ${({u_{1}}^{2})}^{- 1}$ .

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Step 6.2.1

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

$- \frac{69}{2} \int {u_{1}}^{2 \cdot - 1} d u_{1} + \int \frac{97}{4 (x + 3)} d x + \int - \frac{81}{4 (x + 5)} d x$

Step 6.2.2

Multiply $2$ by $- 1$ .

$- \frac{69}{2} \int {u_{1}}^{- 2} d u_{1} + \int \frac{97}{4 (x + 3)} d x + \int - \frac{81}{4 (x + 5)} d x$

Step 7

By the Power Rule, the integral of ${u_{1}}^{- 2}$ with respect to $u_{1}$ is $- {u_{1}}^{- 1}$ .

$- \frac{69}{2} (- {u_{1}}^{- 1} + C) + \int \frac{97}{4 (x + 3)} d x + \int - \frac{81}{4 (x + 5)} d x$

Step 8

Since $\frac{97}{4}$ is constant with respect to $x$ , move $\frac{97}{4}$ out of the integral.

$- \frac{69}{2} (- {u_{1}}^{- 1} + C) + \frac{97}{4} \int \frac{1}{x + 3} d x + \int - \frac{81}{4 (x + 5)} d x$

Step 9

Let $u_{2} = x + 3$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 9.1

Let $u_{2} = x + 3$ . Find $\frac{d u_{2}}{d x}$ .

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Step 9.1.1

Differentiate $x + 3$ .

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

Step 9.1.2

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 9.1.3

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

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

Step 9.1.4

Since $3$ is constant with respect to $x$ , the derivative of $3$ 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 $d u_{2}$ .

$- \frac{69}{2} (- {u_{1}}^{- 1} + C) + \frac{97}{4} \int \frac{1}{u_{2}} d u_{2} + \int - \frac{81}{4 (x + 5)} d x$

Step 10

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

$- \frac{69}{2} (- {u_{1}}^{- 1} + C) + \frac{97}{4} (\ln (| u_{2} |) + C) + \int - \frac{81}{4 (x + 5)} d x$

Step 11

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

$- \frac{69}{2} (- {u_{1}}^{- 1} + C) + \frac{97}{4} (\ln (| u_{2} |) + C) - \int \frac{81}{4 (x + 5)} d x$

Step 12

Since $\frac{81}{4}$ is constant with respect to $x$ , move $\frac{81}{4}$ out of the integral.

$- \frac{69}{2} (- {u_{1}}^{- 1} + C) + \frac{97}{4} (\ln (| u_{2} |) + C) - (\frac{81}{4} \int \frac{1}{x + 5} d x)$

Step 13

Let $u_{3} = x + 5$ . Then $d u_{3} = d x$ . Rewrite using $u_{3}$ and $d$ $u_{3}$ .

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Step 13.1

Let $u_{3} = x + 5$ . Find $\frac{d u_{3}}{d x}$ .

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Step 13.1.1

Differentiate $x + 5$ .

$\frac{d}{d x} [x + 5]$

Step 13.1.2

By the Sum Rule, the derivative of $x + 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [5]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [5]$

Step 13.1.3

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

$1 + \frac{d}{d x} [5]$

Step 13.1.4

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$1 + 0$

Step 13.1.5

Add $1$ and $0$ .

$1$

Step 13.2

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

$- \frac{69}{2} (- {u_{1}}^{- 1} + C) + \frac{97}{4} (\ln (| u_{2} |) + C) - \frac{81}{4} \int \frac{1}{u_{3}} d u_{3}$

Step 14

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

$- \frac{69}{2} (- {u_{1}}^{- 1} + C) + \frac{97}{4} (\ln (| u_{2} |) + C) - \frac{81}{4} (\ln (| u_{3} |) + C)$

Step 15

Simplify.

$\frac{69}{2 u_{1}} + \frac{97 \ln (| u_{2} |)}{4} - \frac{81}{4} \ln (| u_{3} |) + C$

Step 16

Substitute back in for each integration substitution variable.

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Step 16.1

Replace all occurrences of $u_{1}$ with $x + 3$ .

$\frac{69}{2 (x + 3)} + \frac{97 \ln (| u_{2} |)}{4} - \frac{81}{4} \ln (| u_{3} |) + C$

Step 16.2

Replace all occurrences of $u_{2}$ with $x + 3$ .

$\frac{69}{2 (x + 3)} + \frac{97 \ln (| x + 3 |)}{4} - \frac{81}{4} \ln (| u_{3} |) + C$

Step 16.3

Replace all occurrences of $u_{3}$ with $x + 5$ .

$\frac{69}{2 (x + 3)} + \frac{97 \ln (| x + 3 |)}{4} - \frac{81}{4} \ln (| x + 5 |) + C$

Step 17

Reorder terms.

$\frac{69}{2 (x + 3)} + \frac{97}{4} \ln (| x + 3 |) - \frac{81}{4} \ln (| x + 5 |) + C$