Calculus Examples

$\int \frac{y^{3} - y - 1}{(y - 1) (y^{2} - y)} d y$

Step 1

Apply the distributive property.

$\int \frac{y^{3} - y - 1}{y (y^{2} - y) - 1 (y^{2} - y)} d y$

Step 2

Apply the distributive property.

$\int \frac{y^{3} - y - 1}{y \cdot y^{2} + y (- y) - 1 (y^{2} - y)} d y$

Step 3

Apply the distributive property.

$\int \frac{y^{3} - y - 1}{y \cdot y^{2} + y (- y) - 1 y^{2} - 1 (- y)} d y$

Step 4

Reorder $y$ and $- 1$ .

$\int \frac{y^{3} - y - 1}{y \cdot y^{2} - 1 \cdot y \cdot y - 1 y^{2} - 1 (- y)} d y$

Step 5

Raise $y$ to the power of $1$ .

$\int \frac{y^{3} - y - 1}{y^{1} y^{2} - 1 y \cdot y - 1 y^{2} - 1 \cdot - 1 y} d y$

Step 6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int \frac{y^{3} - y - 1}{y^{1 + 2} - 1 y \cdot y - 1 y^{2} - 1 \cdot - 1 y} d y$

Step 7

Add $1$ and $2$ .

$\int \frac{y^{3} - y - 1}{y^{3} - 1 y \cdot y - 1 y^{2} - 1 \cdot - 1 y} d y$

Step 8

Factor out negative.

$\int \frac{y^{3} - y - 1}{y^{3} - (y \cdot y) - 1 y^{2} - 1 \cdot - 1 y} d y$

Step 9

Raise $y$ to the power of $1$ .

$\int \frac{y^{3} - y - 1}{y^{3} - (y^{1} y) - 1 y^{2} - 1 \cdot - 1 y} d y$

Step 10

Raise $y$ to the power of $1$ .

$\int \frac{y^{3} - y - 1}{y^{3} - (y^{1} y^{1}) - 1 y^{2} - 1 \cdot - 1 y} d y$

Step 11

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int \frac{y^{3} - y - 1}{y^{3} - y^{1 + 1} - 1 y^{2} - 1 \cdot - 1 y} d y$

Step 12

Simplify the expression.

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Step 12.1

Add $1$ and $1$ .

$\int \frac{y^{3} - y - 1}{y^{3} - y^{2} - 1 y^{2} - 1 \cdot - 1 y} d y$

Step 12.2

Multiply $- 1$ by $- 1$ .

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

Step 12.3

Multiply $y$ by $1$ .

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

Step 13

Subtract $1 y^{2}$ from $- y^{2}$ .

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

Step 14

Divide $y^{3} - y - 1$ by $y^{3} - 2 y^{2} + y$ .

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Step 14.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$y^{3}$

$2 y^{2}$

$y$

$0$

$y^{3}$

$0 y^{2}$

$y$

$1$

Step 14.2

Divide the highest order term in the dividend $y^{3}$ by the highest order term in divisor $y^{3}$ .

									$1$
	$y^{3}$	-	$2 y^{2}$	+	$y$	+	$0$		$y^{3}$	+	$0 y^{2}$	-	$y$	-	$1$

Step 14.3

Multiply the new quotient term by the divisor.

$1$

$y^{3}$

$2 y^{2}$

$y$

$0$

$y^{3}$

$0 y^{2}$

$y$

$1$

$y^{3}$

$2 y^{2}$

$y$

$0$

Step 14.4

The expression needs to be subtracted from the dividend, so change all the signs in $y^{3} - 2 y^{2} + y + 0$

								$1$
$y^{3}$	-	$2 y^{2}$	+	$y$	+	$0$		$y^{3}$	+	$0 y^{2}$	-	$y$	-	$1$
							-	$y^{3}$	+	$2 y^{2}$	-	$y$	-	$0$

Step 14.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

								$1$
$y^{3}$	-	$2 y^{2}$	+	$y$	+	$0$		$y^{3}$	+	$0 y^{2}$	-	$y$	-	$1$
							-	$y^{3}$	+	$2 y^{2}$	-	$y$	-	$0$
									+	$2 y^{2}$	-	$2 y$	-	$1$

Step 14.6

The final answer is the quotient plus the remainder over the divisor.

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

Step 15

Split the single integral into multiple integrals.

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

Step 16

Apply the constant rule.

$y + C + \int \frac{2 y^{2} - 2 y - 1}{y^{3} - 2 y^{2} + y} d y$

Step 17

Write the fraction using partial fraction decomposition.

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Step 17.1

Decompose the fraction and multiply through by the common denominator.

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Step 17.1.1

Factor the fraction.

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Step 17.1.1.1

Factor $y$ out of $y^{3} - 2 y^{2} + y$ .

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Step 17.1.1.1.1

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

$\frac{2 y^{2} - 2 y - 1}{y \cdot y^{2} - 2 y^{2} + y}$

Step 17.1.1.1.2

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

$\frac{2 y^{2} - 2 y - 1}{y \cdot y^{2} + y (- 2 y) + y}$

Step 17.1.1.1.3

Raise $y$ to the power of $1$ .

$\frac{2 y^{2} - 2 y - 1}{y \cdot y^{2} + y (- 2 y) + y^{1}}$

Step 17.1.1.1.4

Factor $y$ out of $y^{1}$ .

$\frac{2 y^{2} - 2 y - 1}{y \cdot y^{2} + y (- 2 y) + y \cdot 1}$

Step 17.1.1.1.5

Factor $y$ out of $y \cdot y^{2} + y (- 2 y)$ .

$\frac{2 y^{2} - 2 y - 1}{y (y^{2} - 2 y) + y \cdot 1}$

Step 17.1.1.1.6

Factor $y$ out of $y (y^{2} - 2 y) + y \cdot 1$ .

$\frac{2 y^{2} - 2 y - 1}{y (y^{2} - 2 y + 1)}$

Step 17.1.1.2

Factor using the perfect square rule.

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Step 17.1.1.2.1

Rewrite $1$ as $1^{2}$ .

$\frac{2 y^{2} - 2 y - 1}{y (y^{2} - 2 y + 1^{2})}$

Step 17.1.1.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 y = 2 \cdot y \cdot 1$

Step 17.1.1.2.3

Rewrite the polynomial.

$\frac{2 y^{2} - 2 y - 1}{y (y^{2} - 2 \cdot y \cdot 1 + 1^{2})}$

Step 17.1.1.2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = y$ and $b = 1$ .

$\frac{2 y^{2} - 2 y - 1}{y {(y - 1)}^{2}}$

Step 17.1.2

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $B$ .

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

Step 17.1.3

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

Step 17.1.4

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

$\frac{(2 y^{2} - 2 y - 1) (y {(y - 1)}^{2})}{y {(y - 1)}^{2}} = \frac{A (y {(y - 1)}^{2})}{y} + \frac{(B) (y {(y - 1)}^{2})}{{(y - 1)}^{2}} + \frac{(C) (y {(y - 1)}^{2})}{y - 1}$

Step 17.1.5

Cancel the common factor of $y$ .

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Step 17.1.5.1

Cancel the common factor.

$\frac{(2 y^{2} - 2 y - 1) (y {(y - 1)}^{2})}{y {(y - 1)}^{2}} = \frac{A (y {(y - 1)}^{2})}{y} + \frac{(B) (y {(y - 1)}^{2})}{{(y - 1)}^{2}} + \frac{(C) (y {(y - 1)}^{2})}{y - 1}$

Step 17.1.5.2

Rewrite the expression.

$\frac{(2 y^{2} - 2 y - 1) ({(y - 1)}^{2})}{{(y - 1)}^{2}} = \frac{A (y {(y - 1)}^{2})}{y} + \frac{(B) (y {(y - 1)}^{2})}{{(y - 1)}^{2}} + \frac{(C) (y {(y - 1)}^{2})}{y - 1}$

Step 17.1.6

Cancel the common factor of ${(y - 1)}^{2}$ .

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Step 17.1.6.1

Cancel the common factor.

$\frac{(2 y^{2} - 2 y - 1) {(y - 1)}^{2}}{{(y - 1)}^{2}} = \frac{A (y {(y - 1)}^{2})}{y} + \frac{(B) (y {(y - 1)}^{2})}{{(y - 1)}^{2}} + \frac{(C) (y {(y - 1)}^{2})}{y - 1}$

Step 17.1.6.2

Divide $2 y^{2} - 2 y - 1$ by $1$ .

$2 y^{2} - 2 y - 1 = \frac{A (y {(y - 1)}^{2})}{y} + \frac{(B) (y {(y - 1)}^{2})}{{(y - 1)}^{2}} + \frac{(C) (y {(y - 1)}^{2})}{y - 1}$

Step 17.1.7

Simplify each term.

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Step 17.1.7.1

Cancel the common factor of $y$ .

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Step 17.1.7.1.1

Cancel the common factor.

$2 y^{2} - 2 y - 1 = \frac{A (y {(y - 1)}^{2})}{y} + \frac{(B) (y {(y - 1)}^{2})}{{(y - 1)}^{2}} + \frac{(C) (y {(y - 1)}^{2})}{y - 1}$

Step 17.1.7.1.2

Divide $A ({(y - 1)}^{2})$ by $1$ .

$2 y^{2} - 2 y - 1 = A ({(y - 1)}^{2}) + \frac{(B) (y {(y - 1)}^{2})}{{(y - 1)}^{2}} + \frac{(C) (y {(y - 1)}^{2})}{y - 1}$

Step 17.1.7.2

Rewrite ${(y - 1)}^{2}$ as $(y - 1) (y - 1)$ .

$2 y^{2} - 2 y - 1 = A ((y - 1) (y - 1)) + \frac{(B) (y {(y - 1)}^{2})}{{(y - 1)}^{2}} + \frac{(C) (y {(y - 1)}^{2})}{y - 1}$

Step 17.1.7.3

Expand $(y - 1) (y - 1)$ using the FOIL Method.

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Step 17.1.7.3.1

Apply the distributive property.

$2 y^{2} - 2 y - 1 = A (y (y - 1) - 1 (y - 1)) + \frac{(B) (y {(y - 1)}^{2})}{{(y - 1)}^{2}} + \frac{(C) (y {(y - 1)}^{2})}{y - 1}$

Step 17.1.7.3.2

Apply the distributive property.

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

Step 17.1.7.3.3

Apply the distributive property.

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

Step 17.1.7.4

Simplify and combine like terms.

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Step 17.1.7.4.1

Simplify each term.

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Step 17.1.7.4.1.1

Multiply $y$ by $y$ .

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

Step 17.1.7.4.1.2

Move $- 1$ to the left of $y$ .

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

Step 17.1.7.4.1.3

Rewrite $- 1 y$ as $- y$ .

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

Step 17.1.7.4.1.4

Rewrite $- 1 y$ as $- y$ .

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

Step 17.1.7.4.1.5

Multiply $- 1$ by $- 1$ .

$2 y^{2} - 2 y - 1 = A (y^{2} - y - y + 1) + \frac{(B) (y {(y - 1)}^{2})}{{(y - 1)}^{2}} + \frac{(C) (y {(y - 1)}^{2})}{y - 1}$

Step 17.1.7.4.2

Subtract $y$ from $- y$ .

$2 y^{2} - 2 y - 1 = A (y^{2} - 2 y + 1) + \frac{(B) (y {(y - 1)}^{2})}{{(y - 1)}^{2}} + \frac{(C) (y {(y - 1)}^{2})}{y - 1}$

Step 17.1.7.5

Apply the distributive property.

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

Step 17.1.7.6

Simplify.

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Step 17.1.7.6.1

Rewrite using the commutative property of multiplication.

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

Step 17.1.7.6.2

Multiply $A$ by $1$ .

$2 y^{2} - 2 y - 1 = A y^{2} - 2 A y + A + \frac{(B) (y {(y - 1)}^{2})}{{(y - 1)}^{2}} + \frac{(C) (y {(y - 1)}^{2})}{y - 1}$

Step 17.1.7.7

Cancel the common factor of ${(y - 1)}^{2}$ .

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Step 17.1.7.7.1

Cancel the common factor.

$2 y^{2} - 2 y - 1 = A y^{2} - 2 A y + A + \frac{B (y {(y - 1)}^{2})}{{(y - 1)}^{2}} + \frac{(C) (y {(y - 1)}^{2})}{y - 1}$

Step 17.1.7.7.2

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

$2 y^{2} - 2 y - 1 = A y^{2} - 2 A y + A + (B) (y) + \frac{(C) (y {(y - 1)}^{2})}{y - 1}$

Step 17.1.7.8

Cancel the common factor of ${(y - 1)}^{2}$ and $y - 1$ .

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Step 17.1.7.8.1

Factor $y - 1$ out of $(C) (y {(y - 1)}^{2})$ .

$2 y^{2} - 2 y - 1 = A y^{2} - 2 A y + A + B y + \frac{(y - 1) (C (y (y - 1)))}{y - 1}$

Step 17.1.7.8.2

Cancel the common factors.

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Step 17.1.7.8.2.1

Multiply by $1$ .

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

Step 17.1.7.8.2.2

Cancel the common factor.

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

Step 17.1.7.8.2.3

Rewrite the expression.

$2 y^{2} - 2 y - 1 = A y^{2} - 2 A y + A + B y + \frac{C (y (y - 1))}{1}$

Step 17.1.7.8.2.4

Divide $C (y (y - 1))$ by $1$ .

$2 y^{2} - 2 y - 1 = A y^{2} - 2 A y + A + B y + C (y (y - 1))$

Step 17.1.7.9

Apply the distributive property.

$2 y^{2} - 2 y - 1 = A y^{2} - 2 A y + A + B y + C (y \cdot y + y \cdot - 1)$

Step 17.1.7.10

Multiply $y$ by $y$ .

$2 y^{2} - 2 y - 1 = A y^{2} - 2 A y + A + B y + C (y^{2} + y \cdot - 1)$

Step 17.1.7.11

Move $- 1$ to the left of $y$ .

$2 y^{2} - 2 y - 1 = A y^{2} - 2 A y + A + B y + C (y^{2} - 1 \cdot y)$

Step 17.1.7.12

Rewrite $- 1 y$ as $- y$ .

$2 y^{2} - 2 y - 1 = A y^{2} - 2 A y + A + B y + C (y^{2} - y)$

Step 17.1.7.13

Apply the distributive property.

$2 y^{2} - 2 y - 1 = A y^{2} - 2 A y + A + B y + C y^{2} + C (- y)$

Step 17.1.7.14

Rewrite using the commutative property of multiplication.

$2 y^{2} - 2 y - 1 = A y^{2} - 2 A y + A + B y + C y^{2} - C y$

Step 17.1.8

Simplify the expression.

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Step 17.1.8.1

Move $A$ .

$2 y^{2} - 2 y - 1 = A y^{2} - 2 y A + A + B y + C y^{2} - C y$

Step 17.1.8.2

Reorder $B$ and $y$ .

$2 y^{2} - 2 y - 1 = A y^{2} - 2 y A + A + B y + C y^{2} - C y$

Step 17.1.8.3

Move $A$ .

$2 y^{2} - 2 y - 1 = A y^{2} - 2 y A + B y + C y^{2} - C y + A$

Step 17.1.8.4

Move $B y$ .

$2 y^{2} - 2 y - 1 = A y^{2} - 2 y A + C y^{2} + B y - C y + A$

Step 17.1.8.5

Move $- 2 y A$ .

$2 y^{2} - 2 y - 1 = A y^{2} + C y^{2} - 2 y A + B y - C y + A$

Step 17.2

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

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Step 17.2.1

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.

$2 = A + C$

Step 17.2.2

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.

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

Step 17.2.3

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.

$- 1 = A$

Step 17.2.4

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

$2 = A + C$

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

$- 1 = A$

$2 = A + C$

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

$- 1 = A$

Step 17.3

Solve the system of equations.

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Step 17.3.1

Rewrite the equation as $A = - 1$ .

$A = - 1$

$2 = A + C$

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

Step 17.3.2

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

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Step 17.3.2.1

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

$2 = (- 1) + C$

$A = - 1$

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

Step 17.3.2.2

Simplify the right side.

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Step 17.3.2.2.1

Remove parentheses.

$2 = - 1 + C$

$A = - 1$

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

$2 = - 1 + C$

$A = - 1$

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

Step 17.3.2.3

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

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

$2 = - 1 + C$

$A = - 1$

Step 17.3.2.4

Simplify the right side.

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Step 17.3.2.4.1

Simplify each term.

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Step 17.3.2.4.1.1

Multiply $- 2$ by $- 1$ .

$- 2 = 2 + B - 1 C$

$2 = - 1 + C$

$A = - 1$

Step 17.3.2.4.1.2

Rewrite $- 1 C$ as $- C$ .

$- 2 = 2 + B - C$

$2 = - 1 + C$

$A = - 1$

$- 2 = 2 + B - C$

$2 = - 1 + C$

$A = - 1$

$- 2 = 2 + B - C$

$2 = - 1 + C$

$A = - 1$

$- 2 = 2 + B - C$

$2 = - 1 + C$

$A = - 1$

Step 17.3.3

Solve for $C$ in $2 = - 1 + C$ .

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Step 17.3.3.1

Rewrite the equation as $- 1 + C = 2$ .

$- 1 + C = 2$

$- 2 = 2 + B - C$

$A = - 1$

Step 17.3.3.2

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

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Step 17.3.3.2.1

Add $1$ to both sides of the equation.

$C = 2 + 1$

$- 2 = 2 + B - C$

$A = - 1$

Step 17.3.3.2.2

Add $2$ and $1$ .

$C = 3$

$- 2 = 2 + B - C$

$A = - 1$

$C = 3$

$- 2 = 2 + B - C$

$A = - 1$

$C = 3$

$- 2 = 2 + B - C$

$A = - 1$

Step 17.3.4

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

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Step 17.3.4.1

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

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

$C = 3$

$A = - 1$

Step 17.3.4.2

Simplify the right side.

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Step 17.3.4.2.1

Simplify $2 + B - (3)$ .

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Step 17.3.4.2.1.1

Multiply $- 1$ by $3$ .

$- 2 = 2 + B - 3$

$C = 3$

$A = - 1$

Step 17.3.4.2.1.2

Subtract $3$ from $2$ .

$- 2 = B - 1$

$C = 3$

$A = - 1$

$- 2 = B - 1$

$C = 3$

$A = - 1$

$- 2 = B - 1$

$C = 3$

$A = - 1$

$- 2 = B - 1$

$C = 3$

$A = - 1$

Step 17.3.5

Solve for $B$ in $- 2 = B - 1$ .

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Step 17.3.5.1

Rewrite the equation as $B - 1 = - 2$ .

$B - 1 = - 2$

$C = 3$

$A = - 1$

Step 17.3.5.2

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

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Step 17.3.5.2.1

Add $1$ to both sides of the equation.

$B = - 2 + 1$

$C = 3$

$A = - 1$

Step 17.3.5.2.2

Add $- 2$ and $1$ .

$B = - 1$

$C = 3$

$A = - 1$

$B = - 1$

$C = 3$

$A = - 1$

$B = - 1$

$C = 3$

$A = - 1$

Step 17.3.6

Solve the system of equations.

$B = - 1 C = 3 A = - 1$

Step 17.3.7

List all of the solutions.

$B = - 1, C = 3, A = - 1$

Step 17.4

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

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

Step 17.5

Simplify.

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Step 17.5.1

Move the negative in front of the fraction.

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

Step 17.5.2

Move the negative in front of the fraction.

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

Step 18

Split the single integral into multiple integrals.

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

Step 19

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

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

Step 20

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

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

Step 21

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

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

Step 22

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

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Step 22.1

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

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Step 22.1.1

Differentiate $y - 1$ .

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

Step 22.1.2

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

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

Step 22.1.3

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

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

Step 22.1.4

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

$1 + 0$

Step 22.1.5

Add $1$ and $0$ .

$1$

Step 22.2

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

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

Step 23

Apply basic rules of exponents.

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Step 23.1

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

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

Step 23.2

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

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Step 23.2.1

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

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

Step 23.2.2

Multiply $2$ by $- 1$ .

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

Step 24

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

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

Step 25

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

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

Step 26

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

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Step 26.1

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

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Step 26.1.1

Differentiate $y - 1$ .

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

Step 26.1.2

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

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

Step 26.1.3

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

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

Step 26.1.4

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

$1 + 0$

Step 26.1.5

Add $1$ and $0$ .

$1$

Step 26.2

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

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

Step 27

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

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

Step 28

Simplify.

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

Step 29

Substitute back in for each integration substitution variable.

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Step 29.1

Replace all occurrences of $u_{1}$ with $y - 1$ .

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

Step 29.2

Replace all occurrences of $u_{2}$ with $y - 1$ .

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