Calculus Examples

$\int_{1}^{2} \frac{4 y^{2} - 7 y - 12}{y (y + 2) (y - 3)} d y$

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 $B$ .

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

Step 1.1.2

$\frac{A}{y} + \frac{B}{y + 2} + \frac{C}{y - 3}$

Step 1.1.3

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

$\frac{(4 y^{2} - 7 y - 12) (y (y + 2) (y - 3))}{y (y + 2) (y - 3)} = \frac{A (y (y + 2) (y - 3))}{y} + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.4

Cancel the common factor of $y$ .

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Step 1.1.4.1

Cancel the common factor.

$\frac{(4 y^{2} - 7 y - 12) (y (y + 2) (y - 3))}{y (y + 2) (y - 3)} = \frac{A (y (y + 2) (y - 3))}{y} + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.4.2

Rewrite the expression.

$\frac{(4 y^{2} - 7 y - 12) ((y + 2) (y - 3))}{(y + 2) (y - 3)} = \frac{A (y (y + 2) (y - 3))}{y} + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.5

Cancel the common factor of $y + 2$ .

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Step 1.1.5.1

Cancel the common factor.

$\frac{(4 y^{2} - 7 y - 12) ((y + 2) (y - 3))}{(y + 2) (y - 3)} = \frac{A (y (y + 2) (y - 3))}{y} + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.5.2

Rewrite the expression.

$\frac{(4 y^{2} - 7 y - 12) (y - 3)}{y - 3} = \frac{A (y (y + 2) (y - 3))}{y} + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.6

Cancel the common factor of $y - 3$ .

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Step 1.1.6.1

Cancel the common factor.

$\frac{(4 y^{2} - 7 y - 12) (y - 3)}{y - 3} = \frac{A (y (y + 2) (y - 3))}{y} + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.6.2

Divide $4 y^{2} - 7 y - 12$ by $1$ .

$4 y^{2} - 7 y - 12 = \frac{A (y (y + 2) (y - 3))}{y} + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

Cancel the common factor of $y$ .

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Step 1.1.7.1.1

Cancel the common factor.

$4 y^{2} - 7 y - 12 = \frac{A (y (y + 2) (y - 3))}{y} + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.1.2

Divide $A ((y + 2) (y - 3))$ by $1$ .

$4 y^{2} - 7 y - 12 = A ((y + 2) (y - 3)) + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.2

Expand $(y + 2) (y - 3)$ using the FOIL Method.

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Step 1.1.7.2.1

Apply the distributive property.

$4 y^{2} - 7 y - 12 = A (y (y - 3) + 2 (y - 3)) + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.2.2

Apply the distributive property.

$4 y^{2} - 7 y - 12 = A (y \cdot y + y \cdot - 3 + 2 (y - 3)) + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.2.3

Apply the distributive property.

$4 y^{2} - 7 y - 12 = A (y \cdot y + y \cdot - 3 + 2 y + 2 \cdot - 3) + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.3

Simplify and combine like terms.

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Step 1.1.7.3.1

Simplify each term.

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Step 1.1.7.3.1.1

Multiply $y$ by $y$ .

$4 y^{2} - 7 y - 12 = A (y^{2} + y \cdot - 3 + 2 y + 2 \cdot - 3) + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.3.1.2

Move $- 3$ to the left of $y$ .

$4 y^{2} - 7 y - 12 = A (y^{2} - 3 \cdot y + 2 y + 2 \cdot - 3) + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.3.1.3

Multiply $2$ by $- 3$ .

$4 y^{2} - 7 y - 12 = A (y^{2} - 3 y + 2 y - 6) + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.3.2

Add $- 3 y$ and $2 y$ .

$4 y^{2} - 7 y - 12 = A (y^{2} - y - 6) + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.4

Apply the distributive property.

$4 y^{2} - 7 y - 12 = A y^{2} + A (- y) + A \cdot - 6 + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.5

Simplify.

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Step 1.1.7.5.1

Rewrite using the commutative property of multiplication.

$4 y^{2} - 7 y - 12 = A y^{2} - A y + A \cdot - 6 + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.5.2

Move $- 6$ to the left of $A$ .

$4 y^{2} - 7 y - 12 = A y^{2} - A y - 6 \cdot A + \frac{(B) (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.6

Cancel the common factor of $y + 2$ .

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Step 1.1.7.6.1

Cancel the common factor.

$4 y^{2} - 7 y - 12 = A y^{2} - A y - 6 A + \frac{B (y (y + 2) (y - 3))}{y + 2} + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.6.2

Divide $(B) (y (y - 3))$ by $1$ .

$4 y^{2} - 7 y - 12 = A y^{2} - A y - 6 A + (B) (y (y - 3)) + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.7

Apply the distributive property.

$4 y^{2} - 7 y - 12 = A y^{2} - A y - 6 A + B (y \cdot y + y \cdot - 3) + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.8

Multiply $y$ by $y$ .

$4 y^{2} - 7 y - 12 = A y^{2} - A y - 6 A + B (y^{2} + y \cdot - 3) + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.9

Move $- 3$ to the left of $y$ .

$4 y^{2} - 7 y - 12 = A y^{2} - A y - 6 A + B (y^{2} - 3 \cdot y) + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.10

Apply the distributive property.

$4 y^{2} - 7 y - 12 = A y^{2} - A y - 6 A + B y^{2} + B (- 3 y) + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.11

Rewrite using the commutative property of multiplication.

$4 y^{2} - 7 y - 12 = A y^{2} - A y - 6 A + B y^{2} - 3 B y + \frac{(C) (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.12

Cancel the common factor of $y - 3$ .

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Step 1.1.7.12.1

Cancel the common factor.

$4 y^{2} - 7 y - 12 = A y^{2} - A y - 6 A + B y^{2} - 3 B y + \frac{C (y (y + 2) (y - 3))}{y - 3}$

Step 1.1.7.12.2

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

$4 y^{2} - 7 y - 12 = A y^{2} - A y - 6 A + B y^{2} - 3 B y + (C) (y (y + 2))$

Step 1.1.7.13

Apply the distributive property.

$4 y^{2} - 7 y - 12 = A y^{2} - A y - 6 A + B y^{2} - 3 B y + C (y \cdot y + y \cdot 2)$

Step 1.1.7.14

Multiply $y$ by $y$ .

$4 y^{2} - 7 y - 12 = A y^{2} - A y - 6 A + B y^{2} - 3 B y + C (y^{2} + y \cdot 2)$

Step 1.1.7.15

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

$4 y^{2} - 7 y - 12 = A y^{2} - A y - 6 A + B y^{2} - 3 B y + C (y^{2} + 2 \cdot y)$

Step 1.1.7.16

Apply the distributive property.

$4 y^{2} - 7 y - 12 = A y^{2} - A y - 6 A + B y^{2} - 3 B y + C y^{2} + C (2 y)$

Step 1.1.7.17

Rewrite using the commutative property of multiplication.

$4 y^{2} - 7 y - 12 = A y^{2} - A y - 6 A + B y^{2} - 3 B y + C y^{2} + 2 C y$

Step 1.1.8

Simplify the expression.

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Step 1.1.8.1

Move $A$ .

$4 y^{2} - 7 y - 12 = A y^{2} - y A - 6 A + B y^{2} - 3 B y + C y^{2} + 2 C y$

Step 1.1.8.2

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

$4 y^{2} - 7 y - 12 = A y^{2} - y A - 6 A + B y^{2} - 3 B y + C y^{2} + 2 C y$

Step 1.1.8.3

Move $B$ .

$4 y^{2} - 7 y - 12 = A y^{2} - y A - 6 A + B y^{2} - 3 y B + C y^{2} + 2 C y$

Step 1.1.8.4

Move $- 6 A$ .

$4 y^{2} - 7 y - 12 = A y^{2} - y A + B y^{2} - 3 y B + C y^{2} + 2 C y - 6 A$

Step 1.1.8.5

Move $- 3 y B$ .

$4 y^{2} - 7 y - 12 = A y^{2} - y A + B y^{2} + C y^{2} - 3 y B + 2 C y - 6 A$

Step 1.1.8.6

Move $- y A$ .

$4 y^{2} - 7 y - 12 = A y^{2} + B y^{2} + C y^{2} - y A - 3 y B + 2 C y - 6 A$

Step 1.2

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

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Step 1.2.1

Create an equation for the partial fraction variables by equating the coefficients of $y^{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$4 = A + B + C$

Step 1.2.2

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

$- 7 = - 1 A - 3 B + 2 C$

Step 1.2.3

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

$- 12 = - 6 A$

Step 1.2.4

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

$4 = A + B + C$

$- 7 = - 1 A - 3 B + 2 C$

$- 12 = - 6 A$

$4 = A + B + C$

$- 7 = - 1 A - 3 B + 2 C$

$- 12 = - 6 A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $A$ in $- 12 = - 6 A$ .

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Step 1.3.1.1

Rewrite the equation as $- 6 A = - 12$ .

$- 6 A = - 12$

$4 = A + B + C$

$- 7 = - 1 A - 3 B + 2 C$

Step 1.3.1.2

Divide each term in $- 6 A = - 12$ by $- 6$ and simplify.

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Step 1.3.1.2.1

Divide each term in $- 6 A = - 12$ by $- 6$ .

$\frac{- 6 A}{- 6} = \frac{- 12}{- 6}$

$4 = A + B + C$

$- 7 = - 1 A - 3 B + 2 C$

Step 1.3.1.2.2

Simplify the left side.

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Step 1.3.1.2.2.1

Cancel the common factor of $- 6$ .

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Step 1.3.1.2.2.1.1

Cancel the common factor.

$\frac{- 6 A}{- 6} = \frac{- 12}{- 6}$

$4 = A + B + C$

$- 7 = - 1 A - 3 B + 2 C$

Step 1.3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{- 12}{- 6}$

$4 = A + B + C$

$- 7 = - 1 A - 3 B + 2 C$

$A = \frac{- 12}{- 6}$

$4 = A + B + C$

$- 7 = - 1 A - 3 B + 2 C$

$A = \frac{- 12}{- 6}$

$4 = A + B + C$

$- 7 = - 1 A - 3 B + 2 C$

Step 1.3.1.2.3

Simplify the right side.

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Step 1.3.1.2.3.1

Divide $- 12$ by $- 6$ .

$A = 2$

$4 = A + B + C$

$- 7 = - 1 A - 3 B + 2 C$

$A = 2$

$4 = A + B + C$

$- 7 = - 1 A - 3 B + 2 C$

$A = 2$

$4 = A + B + C$

$- 7 = - 1 A - 3 B + 2 C$

$A = 2$

$4 = A + B + C$

$- 7 = - 1 A - 3 B + 2 C$

Step 1.3.2

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

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Step 1.3.2.1

Replace all occurrences of $A$ in $4 = A + B + C$ with $2$ .

$4 = (2) + B + C$

$A = 2$

$- 7 = - 1 A - 3 B + 2 C$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Remove parentheses.

$4 = 2 + B + C$

$A = 2$

$- 7 = - 1 A - 3 B + 2 C$

$4 = 2 + B + C$

$A = 2$

$- 7 = - 1 A - 3 B + 2 C$

Step 1.3.2.3

Replace all occurrences of $A$ in $- 7 = - 1 A - 3 B + 2 C$ with $2$ .

$- 7 = - 1 \cdot 2 - 3 B + 2 C$

$4 = 2 + B + C$

$A = 2$

Step 1.3.2.4

Simplify the right side.

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Step 1.3.2.4.1

Multiply $- 1$ by $2$ .

$- 7 = - 2 - 3 B + 2 C$

$4 = 2 + B + C$

$A = 2$

$- 7 = - 2 - 3 B + 2 C$

$4 = 2 + B + C$

$A = 2$

$- 7 = - 2 - 3 B + 2 C$

$4 = 2 + B + C$

$A = 2$

Step 1.3.3

Solve for $B$ in $4 = 2 + B + C$ .

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Step 1.3.3.1

Rewrite the equation as $2 + B + C = 4$ .

$2 + B + C = 4$

$- 7 = - 2 - 3 B + 2 C$

$A = 2$

Step 1.3.3.2

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

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Step 1.3.3.2.1

Subtract $2$ from both sides of the equation.

$B + C = 4 - 2$

$- 7 = - 2 - 3 B + 2 C$

$A = 2$

Step 1.3.3.2.2

Subtract $C$ from both sides of the equation.

$B = 4 - 2 - C$

$- 7 = - 2 - 3 B + 2 C$

$A = 2$

Step 1.3.3.2.3

Subtract $2$ from $4$ .

$B = 2 - C$

$- 7 = - 2 - 3 B + 2 C$

$A = 2$

$B = 2 - C$

$- 7 = - 2 - 3 B + 2 C$

$A = 2$

$B = 2 - C$

$- 7 = - 2 - 3 B + 2 C$

$A = 2$

Step 1.3.4

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

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Step 1.3.4.1

Replace all occurrences of $B$ in $- 7 = - 2 - 3 B + 2 C$ with $2 - C$ .

$- 7 = - 2 - 3 (2 - C) + 2 C$

$B = 2 - C$

$A = 2$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

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

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Step 1.3.4.2.1.1

Simplify each term.

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Step 1.3.4.2.1.1.1

Apply the distributive property.

$- 7 = - 2 - 3 \cdot 2 - 3 (- C) + 2 C$

$B = 2 - C$

$A = 2$

Step 1.3.4.2.1.1.2

Multiply $- 3$ by $2$ .

$- 7 = - 2 - 6 - 3 (- C) + 2 C$

$B = 2 - C$

$A = 2$

Step 1.3.4.2.1.1.3

Multiply $- 1$ by $- 3$ .

$- 7 = - 2 - 6 + 3 C + 2 C$

$B = 2 - C$

$A = 2$

$- 7 = - 2 - 6 + 3 C + 2 C$

$B = 2 - C$

$A = 2$

Step 1.3.4.2.1.2

Simplify by adding terms.

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Step 1.3.4.2.1.2.1

Subtract $6$ from $- 2$ .

$- 7 = - 8 + 3 C + 2 C$

$B = 2 - C$

$A = 2$

Step 1.3.4.2.1.2.2

Add $3 C$ and $2 C$ .

$- 7 = - 8 + 5 C$

$B = 2 - C$

$A = 2$

$- 7 = - 8 + 5 C$

$B = 2 - C$

$A = 2$

$- 7 = - 8 + 5 C$

$B = 2 - C$

$A = 2$

$- 7 = - 8 + 5 C$

$B = 2 - C$

$A = 2$

$- 7 = - 8 + 5 C$

$B = 2 - C$

$A = 2$

Step 1.3.5

Solve for $C$ in $- 7 = - 8 + 5 C$ .

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Step 1.3.5.1

Rewrite the equation as $- 8 + 5 C = - 7$ .

$- 8 + 5 C = - 7$

$B = 2 - C$

$A = 2$

Step 1.3.5.2

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

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Step 1.3.5.2.1

Add $8$ to both sides of the equation.

$5 C = - 7 + 8$

$B = 2 - C$

$A = 2$

Step 1.3.5.2.2

Add $- 7$ and $8$ .

$5 C = 1$

$B = 2 - C$

$A = 2$

$5 C = 1$

$B = 2 - C$

$A = 2$

Step 1.3.5.3

Divide each term in $5 C = 1$ by $5$ and simplify.

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Step 1.3.5.3.1

Divide each term in $5 C = 1$ by $5$ .

$\frac{5 C}{5} = \frac{1}{5}$

$B = 2 - C$

$A = 2$

Step 1.3.5.3.2

Simplify the left side.

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Step 1.3.5.3.2.1

Cancel the common factor of $5$ .

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Step 1.3.5.3.2.1.1

Cancel the common factor.

$\frac{5 C}{5} = \frac{1}{5}$

$B = 2 - C$

$A = 2$

Step 1.3.5.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{1}{5}$

$B = 2 - C$

$A = 2$

$C = \frac{1}{5}$

$B = 2 - C$

$A = 2$

$C = \frac{1}{5}$

$B = 2 - C$

$A = 2$

$C = \frac{1}{5}$

$B = 2 - C$

$A = 2$

$C = \frac{1}{5}$

$B = 2 - C$

$A = 2$

Step 1.3.6

Replace all occurrences of $C$ with $\frac{1}{5}$ in each equation.

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Step 1.3.6.1

Replace all occurrences of $C$ in $B = 2 - C$ with $\frac{1}{5}$ .

$B = 2 - (\frac{1}{5})$

$C = \frac{1}{5}$

$A = 2$

Step 1.3.6.2

Simplify the right side.

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Step 1.3.6.2.1

Simplify $2 - (\frac{1}{5})$ .

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Step 1.3.6.2.1.1

To write $2$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$B = 2 \cdot \frac{5}{5} - \frac{1}{5}$

$C = \frac{1}{5}$

$A = 2$

Step 1.3.6.2.1.2

Combine $2$ and $\frac{5}{5}$ .

$B = \frac{2 \cdot 5}{5} - \frac{1}{5}$

$C = \frac{1}{5}$

$A = 2$

Step 1.3.6.2.1.3

Combine the numerators over the common denominator.

$B = \frac{2 \cdot 5 - 1}{5}$

$C = \frac{1}{5}$

$A = 2$

Step 1.3.6.2.1.4

Simplify the numerator.

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Step 1.3.6.2.1.4.1

Multiply $2$ by $5$ .

$B = \frac{10 - 1}{5}$

$C = \frac{1}{5}$

$A = 2$

Step 1.3.6.2.1.4.2

Subtract $1$ from $10$ .

$B = \frac{9}{5}$

$C = \frac{1}{5}$

$A = 2$

$B = \frac{9}{5}$

$C = \frac{1}{5}$

$A = 2$

$B = \frac{9}{5}$

$C = \frac{1}{5}$

$A = 2$

$B = \frac{9}{5}$

$C = \frac{1}{5}$

$A = 2$

$B = \frac{9}{5}$

$C = \frac{1}{5}$

$A = 2$

Step 1.3.7

List all of the solutions.

$B = \frac{9}{5}, C = \frac{1}{5}, A = 2$

Step 1.4

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

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

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{2}{y} + \frac{9}{5} \cdot \frac{1}{y + 2} + \frac{\frac{1}{5}}{y - 3}$

Step 1.5.2

Multiply $\frac{9}{5}$ by $\frac{1}{y + 2}$ .

$\frac{2}{y} + \frac{9}{5 (y + 2)} + \frac{\frac{1}{5}}{y - 3}$

Step 1.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{2}{y} + \frac{9}{5 (y + 2)} + \frac{1}{5} \cdot \frac{1}{y - 3}$

Step 1.5.4

Multiply $\frac{1}{5}$ by $\frac{1}{y - 3}$ .

$\int_{1}^{2} \frac{2}{y} + \frac{9}{5 (y + 2)} + \frac{1}{5 (y - 3)} d y$

Step 2

Split the single integral into multiple integrals.

$\int_{1}^{2} \frac{2}{y} d y + \int_{1}^{2} \frac{9}{5 (y + 2)} d y + \int_{1}^{2} \frac{1}{5 (y - 3)} d y$

Step 3

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

$2 \int_{1}^{2} \frac{1}{y} d y + \int_{1}^{2} \frac{9}{5 (y + 2)} d y + \int_{1}^{2} \frac{1}{5 (y - 3)} d y$

Step 4

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

$2 (\ln (| y |)]_{1}^{2}) + \int_{1}^{2} \frac{9}{5 (y + 2)} d y + \int_{1}^{2} \frac{1}{5 (y - 3)} d y$

Step 5

Since $\frac{9}{5}$ is constant with respect to $y$ , move $\frac{9}{5}$ out of the integral.

$2 (\ln (| y |)]_{1}^{2}) + \frac{9}{5} \int_{1}^{2} \frac{1}{y + 2} d y + \int_{1}^{2} \frac{1}{5 (y - 3)} d y$

Step 6

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

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Step 6.1

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

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Step 6.1.1

Differentiate $y + 2$ .

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

Step 6.1.2

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

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

Step 6.1.3

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

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

Step 6.1.4

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

$1 + 0$

Step 6.1.5

Add $1$ and $0$ .

$1$

Step 6.2

Substitute the lower limit in for $y$ in $u_{1} = y + 2$ .

$u_{lower} = 1 + 2$

Step 6.3

Add $1$ and $2$ .

$u_{lower} = 3$

Step 6.4

Substitute the upper limit in for $y$ in $u_{1} = y + 2$ .

$u_{upper} = 2 + 2$

Step 6.5

Add $2$ and $2$ .

$u_{upper} = 4$

Step 6.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 3$

$u_{upper} = 4$

Step 6.7

Rewrite the problem using $u_{1}$ , $d u_{1}$ , and the new limits of integration.

$2 (\ln (| y |)]_{1}^{2}) + \frac{9}{5} \int_{3}^{4} \frac{1}{u_{1}} d u_{1} + \int_{1}^{2} \frac{1}{5 (y - 3)} d y$

Step 7

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

$2 (\ln (| y |)]_{1}^{2}) + \frac{9}{5} \ln (| u_{1} |)]_{3}^{4} + \int_{1}^{2} \frac{1}{5 (y - 3)} d y$

Step 8

Since $\frac{1}{5}$ is constant with respect to $y$ , move $\frac{1}{5}$ out of the integral.

$2 (\ln (| y |)]_{1}^{2}) + \frac{9}{5} \ln (| u_{1} |)]_{3}^{4} + \frac{1}{5} \int_{1}^{2} \frac{1}{y - 3} d y$

Step 9

Let $u_{2} = y - 3$ . Then $d u_{2} = d y$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 9.1

Let $u_{2} = y - 3$ . Find $\frac{d u_{2}}{d y}$ .

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Step 9.1.1

Differentiate $y - 3$ .

$\frac{d}{d y} [y - 3]$

Step 9.1.2

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

$\frac{d}{d y} [y] + \frac{d}{d y} [- 3]$

Step 9.1.3

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

$1 + \frac{d}{d y} [- 3]$

Step 9.1.4

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

$1 + 0$

Step 9.1.5

Add $1$ and $0$ .

$1$

Step 9.2

Substitute the lower limit in for $y$ in $u_{2} = y - 3$ .

$u_{lower} = 1 - 3$

Step 9.3

Subtract $3$ from $1$ .

$u_{lower} = - 2$

Step 9.4

Substitute the upper limit in for $y$ in $u_{2} = y - 3$ .

$u_{upper} = 2 - 3$

Step 9.5

Subtract $3$ from $2$ .

$u_{upper} = - 1$

Step 9.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = - 2$

$u_{upper} = - 1$

Step 9.7

Rewrite the problem using $u_{2}$ , $d u_{2}$ , and the new limits of integration.

$2 (\ln (| y |)]_{1}^{2}) + \frac{9}{5} \ln (| u_{1} |)]_{3}^{4} + \frac{1}{5} \int_{- 2}^{- 1} \frac{1}{u_{2}} d u_{2}$

Step 10

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

$2 (\ln (| y |)]_{1}^{2}) + \frac{9}{5} \ln (| u_{1} |)]_{3}^{4} + \frac{1}{5} \ln (| u_{2} |)]_{- 2}^{- 1}$

Step 11

Substitute and simplify.

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Step 11.1

Evaluate $\ln (| y |)$ at $2$ and at $1$ .

$2 ((\ln (| 2 |)) - \ln (| 1 |)) + \frac{9}{5} \ln (| u_{1} |)]_{3}^{4} + \frac{1}{5} \ln (| u_{2} |)]_{- 2}^{- 1}$

Step 11.2

Evaluate $\ln (| u_{1} |)$ at $4$ and at $3$ .

$2 ((\ln (| 2 |)) - \ln (| 1 |)) + \frac{9}{5} ((\ln (| 4 |)) - \ln (| 3 |)) + \frac{1}{5} \ln (| u_{2} |)]_{- 2}^{- 1}$

Step 11.3

Evaluate $\ln (| u_{2} |)$ at $- 1$ and at $- 2$ .

$2 ((\ln (| 2 |)) - \ln (| 1 |)) + \frac{9}{5} ((\ln (| 4 |)) - \ln (| 3 |)) + \frac{1}{5} ((\ln (| - 1 |)) - \ln (| - 2 |))$

Step 11.4

Remove parentheses.

$2 (\ln (| 2 |) - \ln (| 1 |)) + \frac{9}{5} (\ln (| 4 |) - \ln (| 3 |)) + \frac{1}{5} (\ln (| - 1 |) - \ln (| - 2 |))$

Step 12

Simplify.

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Step 12.1

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$2 \ln (\frac{| 2 |}{| 1 |}) + \frac{9}{5} (\ln (| 4 |) - \ln (| 3 |)) + \frac{1}{5} (\ln (| - 1 |) - \ln (| - 2 |))$

Step 12.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$2 \ln (\frac{| 2 |}{| 1 |}) + \frac{9}{5} \ln (\frac{| 4 |}{| 3 |}) + \frac{1}{5} (\ln (| - 1 |) - \ln (| - 2 |))$

Step 12.3

Combine $\frac{9}{5}$ and $\ln (\frac{| 4 |}{| 3 |})$ .

$2 \ln (\frac{| 2 |}{| 1 |}) + \frac{9 \ln (\frac{| 4 |}{| 3 |})}{5} + \frac{1}{5} (\ln (| - 1 |) - \ln (| - 2 |))$

Step 12.4

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$2 \ln (\frac{| 2 |}{| 1 |}) + \frac{9 \ln (\frac{| 4 |}{| 3 |})}{5} + \frac{1}{5} \ln (\frac{| - 1 |}{| - 2 |})$

Step 12.5

Combine $\frac{1}{5}$ and $\ln (\frac{| - 1 |}{| - 2 |})$ .

$2 \ln (\frac{| 2 |}{| 1 |}) + \frac{9 \ln (\frac{| 4 |}{| 3 |})}{5} + \frac{\ln (\frac{| - 1 |}{| - 2 |})}{5}$

Step 13

Simplify.

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Step 13.1

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$2 \ln (\frac{2}{| 1 |}) + \frac{9 \ln (\frac{| 4 |}{| 3 |})}{5} + \frac{\ln (\frac{| - 1 |}{| - 2 |})}{5}$

Step 13.2

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$2 \ln (\frac{2}{1}) + \frac{9 \ln (\frac{| 4 |}{| 3 |})}{5} + \frac{\ln (\frac{| - 1 |}{| - 2 |})}{5}$

Step 13.3

Divide $2$ by $1$ .

$2 \ln (2) + \frac{9 \ln (\frac{| 4 |}{| 3 |})}{5} + \frac{\ln (\frac{| - 1 |}{| - 2 |})}{5}$

Step 13.4

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$2 \ln (2) + \frac{9 \ln (\frac{4}{| 3 |})}{5} + \frac{\ln (\frac{| - 1 |}{| - 2 |})}{5}$

Step 13.5

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$2 \ln (2) + \frac{9 \ln (\frac{4}{3})}{5} + \frac{\ln (\frac{| - 1 |}{| - 2 |})}{5}$

Step 13.6

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

$2 \ln (2) + \frac{9 \ln (\frac{4}{3})}{5} + \frac{\ln (\frac{1}{| - 2 |})}{5}$

Step 13.7

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

$2 \ln (2) + \frac{9 \ln (\frac{4}{3})}{5} + \frac{\ln (\frac{1}{2})}{5}$

Step 14

The result can be shown in multiple forms.

Exact Form:

$2 \ln (2) + \frac{9 \ln (\frac{4}{3})}{5} + \frac{\ln (\frac{1}{2})}{5}$

Decimal Form:

$1.76549265 \dots$

Step 15