Calculus Examples

$\frac{2 v^{2} + 6 v + 5}{(v + 2) {(v + 1)}^{2}}$

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 $A$ .

$\frac{A}{v + 2}$

Step 1.1.2

$\frac{A}{v + 2} + \frac{B}{{(v + 1)}^{2}}$

Step 1.1.3

$\frac{A}{v + 2} + \frac{B}{{(v + 1)}^{2}} + \frac{C}{v + 1}$

Step 1.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(v + 2) {(v + 1)}^{2}$ .

$\frac{(2 v^{2} + 6 v + 5) (v + 2) {(v + 1)}^{2}}{(v + 2) {(v + 1)}^{2}} = \frac{(A) (v + 2) {(v + 1)}^{2}}{v + 2} + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.5

Cancel the common factor of $v + 2$ .

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Step 1.1.5.1

Cancel the common factor.

Step 1.1.5.2

Rewrite the expression.

$\frac{(2 v^{2} + 6 v + 5) {(v + 1)}^{2}}{{(v + 1)}^{2}} = \frac{(A) (v + 2) {(v + 1)}^{2}}{v + 2} + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.6

Cancel the common factor of ${(v + 1)}^{2}$ .

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Step 1.1.6.1

Cancel the common factor.

$\frac{(2 v^{2} + 6 v + 5) {(v + 1)}^{2}}{{(v + 1)}^{2}} = \frac{(A) (v + 2) {(v + 1)}^{2}}{v + 2} + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.6.2

Divide $2 v^{2} + 6 v + 5$ by $1$ .

$2 v^{2} + 6 v + 5 = \frac{(A) (v + 2) {(v + 1)}^{2}}{v + 2} + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

Cancel the common factor of $v + 2$ .

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Step 1.1.7.1.1

Cancel the common factor.

$2 v^{2} + 6 v + 5 = \frac{A (v + 2) {(v + 1)}^{2}}{v + 2} + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.1.2

Divide $(A) {(v + 1)}^{2}$ by $1$ .

$2 v^{2} + 6 v + 5 = (A) {(v + 1)}^{2} + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.2

Rewrite ${(v + 1)}^{2}$ as $(v + 1) (v + 1)$ .

$2 v^{2} + 6 v + 5 = A ((v + 1) (v + 1)) + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.3

Expand $(v + 1) (v + 1)$ using the FOIL Method.

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Step 1.1.7.3.1

Apply the distributive property.

$2 v^{2} + 6 v + 5 = A (v (v + 1) + 1 (v + 1)) + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.3.2

Apply the distributive property.

$2 v^{2} + 6 v + 5 = A (v \cdot v + v \cdot 1 + 1 (v + 1)) + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.3.3

Apply the distributive property.

$2 v^{2} + 6 v + 5 = A (v \cdot v + v \cdot 1 + 1 v + 1 \cdot 1) + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.4

Simplify and combine like terms.

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Step 1.1.7.4.1

Simplify each term.

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Step 1.1.7.4.1.1

Multiply $v$ by $v$ .

$2 v^{2} + 6 v + 5 = A (v^{2} + v \cdot 1 + 1 v + 1 \cdot 1) + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.4.1.2

Multiply $v$ by $1$ .

$2 v^{2} + 6 v + 5 = A (v^{2} + v + 1 v + 1 \cdot 1) + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.4.1.3

Multiply $v$ by $1$ .

$2 v^{2} + 6 v + 5 = A (v^{2} + v + v + 1 \cdot 1) + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.4.1.4

Multiply $1$ by $1$ .

$2 v^{2} + 6 v + 5 = A (v^{2} + v + v + 1) + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.4.2

Add $v$ and $v$ .

$2 v^{2} + 6 v + 5 = A (v^{2} + 2 v + 1) + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.5

Apply the distributive property.

$2 v^{2} + 6 v + 5 = A v^{2} + A (2 v) + A \cdot 1 + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.6

Simplify.

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Step 1.1.7.6.1

Rewrite using the commutative property of multiplication.

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A \cdot 1 + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.6.2

Multiply $A$ by $1$ .

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.7

Cancel the common factor of ${(v + 1)}^{2}$ .

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Step 1.1.7.7.1

Cancel the common factor.

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + \frac{(B) (v + 2) {(v + 1)}^{2}}{{(v + 1)}^{2}} + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.7.2

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

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + (B) (v + 2) + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.8

Apply the distributive property.

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + B \cdot 2 + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.9

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

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 \cdot B + \frac{(C) (v + 2) {(v + 1)}^{2}}{v + 1}$

Step 1.1.7.10

Cancel the common factor of ${(v + 1)}^{2}$ and $v + 1$ .

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Step 1.1.7.10.1

Factor $v + 1$ out of $(C) (v + 2) {(v + 1)}^{2}$ .

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 B + \frac{(v + 1) ((C) (v + 2) (v + 1))}{v + 1}$

Step 1.1.7.10.2

Cancel the common factors.

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Step 1.1.7.10.2.1

Multiply by $1$ .

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 B + \frac{(v + 1) ((C) (v + 2) (v + 1))}{(v + 1) \cdot 1}$

Step 1.1.7.10.2.2

Cancel the common factor.

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 B + \frac{(v + 1) ((C) (v + 2) (v + 1))}{(v + 1) \cdot 1}$

Step 1.1.7.10.2.3

Rewrite the expression.

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 B + \frac{(C) (v + 2) (v + 1)}{1}$

Step 1.1.7.10.2.4

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

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 B + (C) (v + 2) (v + 1)$

Step 1.1.7.11

Apply the distributive property.

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 B + (C v + C \cdot 2) (v + 1)$

Step 1.1.7.12

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

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 B + (C v + 2 \cdot C) (v + 1)$

Step 1.1.7.13

Expand $(C v + 2 C) (v + 1)$ using the FOIL Method.

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Step 1.1.7.13.1

Apply the distributive property.

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 B + C v (v + 1) + 2 C (v + 1)$

Step 1.1.7.13.2

Apply the distributive property.

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 B + C v \cdot v + C v \cdot 1 + 2 C (v + 1)$

Step 1.1.7.13.3

Apply the distributive property.

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 B + C v \cdot v + C v \cdot 1 + 2 C v + 2 C \cdot 1$

Step 1.1.7.14

Simplify and combine like terms.

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Step 1.1.7.14.1

Simplify each term.

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Step 1.1.7.14.1.1

Multiply $v$ by $v$ by adding the exponents.

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Step 1.1.7.14.1.1.1

Move $v$ .

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 B + C (v \cdot v) + C v \cdot 1 + 2 C v + 2 C \cdot 1$

Step 1.1.7.14.1.1.2

Multiply $v$ by $v$ .

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 B + C v^{2} + C v \cdot 1 + 2 C v + 2 C \cdot 1$

Step 1.1.7.14.1.2

Multiply $C$ by $1$ .

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 B + C v^{2} + C v + 2 C v + 2 C \cdot 1$

Step 1.1.7.14.1.3

Multiply $2$ by $1$ .

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 B + C v^{2} + C v + 2 C v + 2 C$

Step 1.1.7.14.2

Add $C v$ and $2 C v$ .

$2 v^{2} + 6 v + 5 = A v^{2} + 2 A v + A + B v + 2 B + C v^{2} + 3 C v + 2 C$

Step 1.1.8

Simplify the expression.

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Step 1.1.8.1

Move $A$ .

$2 v^{2} + 6 v + 5 = A v^{2} + 2 v A + A + B v + 2 B + C v^{2} + 3 C v + 2 C$

Step 1.1.8.2

Reorder $B$ and $v$ .

$2 v^{2} + 6 v + 5 = A v^{2} + 2 v A + A + B v + 2 B + C v^{2} + 3 C v + 2 C$

Step 1.1.8.3

Move $2 B$ .

$2 v^{2} + 6 v + 5 = A v^{2} + 2 v A + A + B v + C v^{2} + 3 C v + 2 B + 2 C$

Step 1.1.8.4

Move $A$ .

$2 v^{2} + 6 v + 5 = A v^{2} + 2 v A + B v + C v^{2} + 3 C v + A + 2 B + 2 C$

Step 1.1.8.5

Move $B v$ .

$2 v^{2} + 6 v + 5 = A v^{2} + 2 v A + C v^{2} + B v + 3 C v + A + 2 B + 2 C$

Step 1.1.8.6

Move $2 v A$ .

$2 v^{2} + 6 v + 5 = A v^{2} + C v^{2} + 2 v A + B v + 3 C v + A + 2 B + 2 C$

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 $v^{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$2 = A + C$

Step 1.2.2

Create an equation for the partial fraction variables by equating the coefficients of $v$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$6 = 2 A + B + 3 C$

Step 1.2.3

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $v$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$5 = A + 2 B + 2 C$

Step 1.2.4

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

$2 = A + C$

$6 = 2 A + B + 3 C$

$5 = A + 2 B + 2 C$

$2 = A + C$

$6 = 2 A + B + 3 C$

$5 = A + 2 B + 2 C$

Step 1.3

Solve the system of equations.

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Step 1.3.1

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

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Step 1.3.1.1

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

$A + C = 2$

$6 = 2 A + B + 3 C$

$5 = A + 2 B + 2 C$

Step 1.3.1.2

Subtract $C$ from both sides of the equation.

$A = 2 - C$

$6 = 2 A + B + 3 C$

$5 = A + 2 B + 2 C$

$A = 2 - C$

$6 = 2 A + B + 3 C$

$5 = A + 2 B + 2 C$

Step 1.3.2

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

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Step 1.3.2.1

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

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

$A = 2 - C$

$5 = A + 2 B + 2 C$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

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

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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.

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

$A = 2 - C$

$5 = A + 2 B + 2 C$

Step 1.3.2.2.1.1.2

Multiply $2$ by $2$ .

$6 = 4 + 2 (- C) + B + 3 C$

$A = 2 - C$

$5 = A + 2 B + 2 C$

Step 1.3.2.2.1.1.3

Multiply $- 1$ by $2$ .

$6 = 4 - 2 C + B + 3 C$

$A = 2 - C$

$5 = A + 2 B + 2 C$

$6 = 4 - 2 C + B + 3 C$

$A = 2 - C$

$5 = A + 2 B + 2 C$

Step 1.3.2.2.1.2

Add $- 2 C$ and $3 C$ .

$6 = 4 + B + C$

$A = 2 - C$

$5 = A + 2 B + 2 C$

$6 = 4 + B + C$

$A = 2 - C$

$5 = A + 2 B + 2 C$

$6 = 4 + B + C$

$A = 2 - C$

$5 = A + 2 B + 2 C$

Step 1.3.2.3

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

$5 = (2 - C) + 2 B + 2 C$

$6 = 4 + B + C$

$A = 2 - C$

Step 1.3.2.4

Simplify the right side.

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Step 1.3.2.4.1

Add $- C$ and $2 C$ .

$5 = 2 + 2 B + C$

$6 = 4 + B + C$

$A = 2 - C$

$5 = 2 + 2 B + C$

$6 = 4 + B + C$

$A = 2 - C$

$5 = 2 + 2 B + C$

$6 = 4 + B + C$

$A = 2 - C$

Step 1.3.3

Reorder $2$ and $- C$ .

$A = - C + 2$

$5 = 2 + 2 B + C$

$6 = 4 + B + C$

Step 1.3.4

Solve for $C$ in $5 = 2 + 2 B + C$ .

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Step 1.3.4.1

Rewrite the equation as $2 + 2 B + C = 5$ .

$2 + 2 B + C = 5$

$A = - C + 2$

$6 = 4 + B + C$

Step 1.3.4.2

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

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Step 1.3.4.2.1

Subtract $2$ from both sides of the equation.

$2 B + C = 5 - 2$

$A = - C + 2$

$6 = 4 + B + C$

Step 1.3.4.2.2

Subtract $2 B$ from both sides of the equation.

$C = 5 - 2 - 2 B$

$A = - C + 2$

$6 = 4 + B + C$

Step 1.3.4.2.3

Subtract $2$ from $5$ .

$C = 3 - 2 B$

$A = - C + 2$

$6 = 4 + B + C$

$C = 3 - 2 B$

$A = - C + 2$

$6 = 4 + B + C$

$C = 3 - 2 B$

$A = - C + 2$

$6 = 4 + B + C$

Step 1.3.5

Replace all occurrences of $C$ with $3 - 2 B$ in each equation.

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Step 1.3.5.1

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

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

$C = 3 - 2 B$

$6 = 4 + B + C$

Step 1.3.5.2

Simplify the right side.

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Step 1.3.5.2.1

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

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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.

$A = - 1 \cdot 3 - (- 2 B) + 2$

$C = 3 - 2 B$

$6 = 4 + B + C$

Step 1.3.5.2.1.1.2

Multiply $- 1$ by $3$ .

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

$C = 3 - 2 B$

$6 = 4 + B + C$

Step 1.3.5.2.1.1.3

Multiply $- 2$ by $- 1$ .

$A = - 3 + 2 B + 2$

$C = 3 - 2 B$

$6 = 4 + B + C$

$A = - 3 + 2 B + 2$

$C = 3 - 2 B$

$6 = 4 + B + C$

Step 1.3.5.2.1.2

Add $- 3$ and $2$ .

$A = 2 B - 1$

$C = 3 - 2 B$

$6 = 4 + B + C$

$A = 2 B - 1$

$C = 3 - 2 B$

$6 = 4 + B + C$

$A = 2 B - 1$

$C = 3 - 2 B$

$6 = 4 + B + C$

Step 1.3.5.3

Replace all occurrences of $C$ in $6 = 4 + B + C$ with $3 - 2 B$ .

$6 = 4 + B + 3 - 2 B$

$A = 2 B - 1$

$C = 3 - 2 B$

Step 1.3.5.4

Simplify $6 = 4 + B + 3 - 2 B$ .

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Step 1.3.5.4.1

Simplify the left side.

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Step 1.3.5.4.1.1

Remove parentheses.

$6 = 4 + B + 3 - 2 B$

$A = 2 B - 1$

$C = 3 - 2 B$

$6 = 4 + B + 3 - 2 B$

$A = 2 B - 1$

$C = 3 - 2 B$

Step 1.3.5.4.2

Simplify the right side.

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Step 1.3.5.4.2.1

Simplify $4 + B + 3 - 2 B$ .

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Step 1.3.5.4.2.1.1

Add $4$ and $3$ .

$6 = B + 7 - 2 B$

$A = 2 B - 1$

$C = 3 - 2 B$

Step 1.3.5.4.2.1.2

Subtract $2 B$ from $B$ .

$6 = - B + 7$

$A = 2 B - 1$

$C = 3 - 2 B$

$6 = - B + 7$

$A = 2 B - 1$

$C = 3 - 2 B$

$6 = - B + 7$

$A = 2 B - 1$

$C = 3 - 2 B$

$6 = - B + 7$

$A = 2 B - 1$

$C = 3 - 2 B$

$6 = - B + 7$

$A = 2 B - 1$

$C = 3 - 2 B$

Step 1.3.6

Solve for $B$ in $6 = - B + 7$ .

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Step 1.3.6.1

Rewrite the equation as $- B + 7 = 6$ .

$- B + 7 = 6$

$A = 2 B - 1$

$C = 3 - 2 B$

Step 1.3.6.2

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

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Step 1.3.6.2.1

Subtract $7$ from both sides of the equation.

$- B = 6 - 7$

$A = 2 B - 1$

$C = 3 - 2 B$

Step 1.3.6.2.2

Subtract $7$ from $6$ .

$- B = - 1$

$A = 2 B - 1$

$C = 3 - 2 B$

$- B = - 1$

$A = 2 B - 1$

$C = 3 - 2 B$

Step 1.3.6.3

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

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Step 1.3.6.3.1

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

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

$A = 2 B - 1$

$C = 3 - 2 B$

Step 1.3.6.3.2

Simplify the left side.

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Step 1.3.6.3.2.1

Dividing two negative values results in a positive value.

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

$A = 2 B - 1$

$C = 3 - 2 B$

Step 1.3.6.3.2.2

Divide $B$ by $1$ .

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

$A = 2 B - 1$

$C = 3 - 2 B$

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

$A = 2 B - 1$

$C = 3 - 2 B$

Step 1.3.6.3.3

Simplify the right side.

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Step 1.3.6.3.3.1

Divide $- 1$ by $- 1$ .

$B = 1$

$A = 2 B - 1$

$C = 3 - 2 B$

$B = 1$

$A = 2 B - 1$

$C = 3 - 2 B$

$B = 1$

$A = 2 B - 1$

$C = 3 - 2 B$

$B = 1$

$A = 2 B - 1$

$C = 3 - 2 B$

Step 1.3.7

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

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Step 1.3.7.1

Replace all occurrences of $B$ in $A = 2 B - 1$ with $1$ .

$A = 2 (1) - 1$

$B = 1$

$C = 3 - 2 B$

Step 1.3.7.2

Simplify the right side.

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Step 1.3.7.2.1

Simplify $2 (1) - 1$ .

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Step 1.3.7.2.1.1

Multiply $2$ by $1$ .

$A = 2 - 1$

$B = 1$

$C = 3 - 2 B$

Step 1.3.7.2.1.2

Subtract $1$ from $2$ .

$A = 1$

$B = 1$

$C = 3 - 2 B$

$A = 1$

$B = 1$

$C = 3 - 2 B$

$A = 1$

$B = 1$

$C = 3 - 2 B$

Step 1.3.7.3

Replace all occurrences of $B$ in $C = 3 - 2 B$ with $1$ .

$C = 3 - 2 \cdot 1$

$A = 1$

$B = 1$

Step 1.3.7.4

Simplify the right side.

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Step 1.3.7.4.1

Simplify $3 - 2 \cdot 1$ .

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Step 1.3.7.4.1.1

Multiply $- 2$ by $1$ .

$C = 3 - 2$

$A = 1$

$B = 1$

Step 1.3.7.4.1.2

Subtract $2$ from $3$ .

$C = 1$

$A = 1$

$B = 1$

$C = 1$

$A = 1$

$B = 1$

$C = 1$

$A = 1$

$B = 1$

$C = 1$

$A = 1$

$B = 1$

Step 1.3.8

List all of the solutions.

$C = 1, A = 1, B = 1$

Step 1.4

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

$\frac{1}{v + 2} + \frac{1}{{(v + 1)}^{2}} + \frac{1}{v + 1}$

Step 1.5

Remove the zero from the expression.

$\int \frac{1}{v + 2} + \frac{1}{{(v + 1)}^{2}} + \frac{1}{v + 1} d v$

Step 2

Split the single integral into multiple integrals.

$\int \frac{1}{v + 2} d v + \int \frac{1}{{(v + 1)}^{2}} d v + \int \frac{1}{v + 1} d v$

Step 3

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

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Step 3.1

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

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Step 3.1.1

Differentiate $v + 2$ .

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

Step 3.1.2

By the Sum Rule, the derivative of $v + 2$ with respect to $v$ is $\frac{d}{d v} [v] + \frac{d}{d v} [2]$ .

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

Step 3.1.3

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

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

Step 3.1.4

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

$1 + 0$

Step 3.1.5

Add $1$ and $0$ .

$1$

Step 3.2

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

$\int \frac{1}{u_{1}} d u_{1} + \int \frac{1}{{(v + 1)}^{2}} d v + \int \frac{1}{v + 1} d v$

Step 4

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

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

Step 5

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

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Step 5.1

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

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Step 5.1.1

Differentiate $v + 1$ .

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

Step 5.1.2

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

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

Step 5.1.3

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

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

Step 5.1.4

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

$1 + 0$

Step 5.1.5

Add $1$ and $0$ .

$1$

Step 5.2

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

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

Step 6

Apply basic rules of exponents.

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Step 6.1

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

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

Step 6.2

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

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Step 6.2.1

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

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

Step 6.2.2

Multiply $2$ by $- 1$ .

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

Step 7

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

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

Step 8

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

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Step 8.1

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

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Step 8.1.1

Differentiate $v + 1$ .

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

Step 8.1.2

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

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

Step 8.1.3

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

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

Step 8.1.4

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

$1 + 0$

Step 8.1.5

Add $1$ and $0$ .

$1$

Step 8.2

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

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

Step 9

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

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

Step 10

Simplify.

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Step 10.1

Simplify.

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

Step 10.2

Simplify.

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Step 10.2.1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 10.2.2

To multiply absolute values, multiply the terms inside each absolute value.

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

Step 11

Substitute back in for each integration substitution variable.

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Step 11.1

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

$- \frac{1}{v + 1} + \ln (| u_{1} \cdot u_{3} |) + C$

Step 11.2

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

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

Step 11.3

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

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