Calculus Examples

$\int \frac{6 x^{2} - 2 x - 1}{4 x^{3} - x} d x$

Step 1

Write the fraction using partial fraction decomposition.

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Step 1.1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1.1

Factor the fraction.

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Step 1.1.1.1

Factor $x$ out of $4 x^{3} - x$ .

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Step 1.1.1.1.1

Factor $x$ out of $4 x^{3}$ .

$\frac{6 x^{2} - 2 x - 1}{x (4 x^{2}) - x}$

Step 1.1.1.1.2

Factor $x$ out of $- x$ .

$\frac{6 x^{2} - 2 x - 1}{x (4 x^{2}) + x \cdot - 1}$

Step 1.1.1.1.3

Factor $x$ out of $x (4 x^{2}) + x \cdot - 1$ .

$\frac{6 x^{2} - 2 x - 1}{x (4 x^{2} - 1)}$

Step 1.1.1.2

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$\frac{6 x^{2} - 2 x - 1}{x ({(2 x)}^{2} - 1)}$

Step 1.1.1.3

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

$\frac{6 x^{2} - 2 x - 1}{x ({(2 x)}^{2} - 1^{2})}$

Step 1.1.1.4

Factor.

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Step 1.1.1.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 1$ .

$\frac{6 x^{2} - 2 x - 1}{x ((2 x + 1) (2 x - 1))}$

Step 1.1.1.4.2

Remove unnecessary parentheses.

$\frac{6 x^{2} - 2 x - 1}{x (2 x + 1) (2 x - 1)}$

Step 1.1.2

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

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

Step 1.1.3

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

Step 1.1.4

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

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

Step 1.1.5

Cancel the common factor of $x$ .

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Step 1.1.5.1

Cancel the common factor.

Step 1.1.5.2

Rewrite the expression.

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

Step 1.1.6

Cancel the common factor of $2 x + 1$ .

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Step 1.1.6.1

Cancel the common factor.

Step 1.1.6.2

Rewrite the expression.

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

Step 1.1.7

Cancel the common factor of $2 x - 1$ .

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Step 1.1.7.1

Cancel the common factor.

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

Step 1.1.7.2

Divide $6 x^{2} - 2 x - 1$ by $1$ .

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

Step 1.1.8

Simplify each term.

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Step 1.1.8.1

Cancel the common factor of $x$ .

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Step 1.1.8.1.1

Cancel the common factor.

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

Step 1.1.8.1.2

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

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

Step 1.1.8.2

Expand $(2 x + 1) (2 x - 1)$ using the FOIL Method.

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Step 1.1.8.2.1

Apply the distributive property.

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

Step 1.1.8.2.2

Apply the distributive property.

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

Step 1.1.8.2.3

Apply the distributive property.

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

Step 1.1.8.3

Combine the opposite terms in $2 x (2 x) + 2 x \cdot - 1 + 1 (2 x) + 1 \cdot - 1$ .

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Step 1.1.8.3.1

Reorder the factors in the terms $2 x \cdot - 1$ and $1 (2 x)$ .

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

Step 1.1.8.3.2

Add $- 1 \cdot 2 x$ and $1 \cdot 2 x$ .

$6 x^{2} - 2 x - 1 = A (2 x (2 x) + 0 + 1 \cdot - 1) + \frac{(B) (x (2 x + 1) (2 x - 1))}{2 x + 1} + \frac{(C) (x (2 x + 1) (2 x - 1))}{2 x - 1}$

Step 1.1.8.3.3

Add $2 x (2 x)$ and $0$ .

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

Step 1.1.8.4

Simplify each term.

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Step 1.1.8.4.1

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.4.2

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

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Step 1.1.8.4.2.1

Move $x$ .

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

Step 1.1.8.4.2.2

Multiply $x$ by $x$ .

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

Step 1.1.8.4.3

Multiply $2$ by $2$ .

$6 x^{2} - 2 x - 1 = A (4 x^{2} + 1 \cdot - 1) + \frac{(B) (x (2 x + 1) (2 x - 1))}{2 x + 1} + \frac{(C) (x (2 x + 1) (2 x - 1))}{2 x - 1}$

Step 1.1.8.4.4

Multiply $- 1$ by $1$ .

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

Step 1.1.8.5

Apply the distributive property.

$6 x^{2} - 2 x - 1 = A (4 x^{2}) + A \cdot - 1 + \frac{(B) (x (2 x + 1) (2 x - 1))}{2 x + 1} + \frac{(C) (x (2 x + 1) (2 x - 1))}{2 x - 1}$

Step 1.1.8.6

Rewrite using the commutative property of multiplication.

$6 x^{2} - 2 x - 1 = 4 A x^{2} + A \cdot - 1 + \frac{(B) (x (2 x + 1) (2 x - 1))}{2 x + 1} + \frac{(C) (x (2 x + 1) (2 x - 1))}{2 x - 1}$

Step 1.1.8.7

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

$6 x^{2} - 2 x - 1 = 4 A x^{2} - 1 \cdot A + \frac{(B) (x (2 x + 1) (2 x - 1))}{2 x + 1} + \frac{(C) (x (2 x + 1) (2 x - 1))}{2 x - 1}$

Step 1.1.8.8

Rewrite $- 1 A$ as $- A$ .

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

Step 1.1.8.9

Cancel the common factor of $2 x + 1$ .

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Step 1.1.8.9.1

Cancel the common factor.

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

Step 1.1.8.9.2

Divide $(B) (x (2 x - 1))$ by $1$ .

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

Step 1.1.8.10

Apply the distributive property.

$6 x^{2} - 2 x - 1 = 4 A x^{2} - A + B (x (2 x) + x \cdot - 1) + \frac{(C) (x (2 x + 1) (2 x - 1))}{2 x - 1}$

Step 1.1.8.11

Rewrite using the commutative property of multiplication.

$6 x^{2} - 2 x - 1 = 4 A x^{2} - A + B (2 x \cdot x + x \cdot - 1) + \frac{(C) (x (2 x + 1) (2 x - 1))}{2 x - 1}$

Step 1.1.8.12

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

$6 x^{2} - 2 x - 1 = 4 A x^{2} - A + B (2 x \cdot x - 1 \cdot x) + \frac{(C) (x (2 x + 1) (2 x - 1))}{2 x - 1}$

Step 1.1.8.13

Simplify each term.

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Step 1.1.8.13.1

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

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Step 1.1.8.13.1.1

Move $x$ .

$6 x^{2} - 2 x - 1 = 4 A x^{2} - A + B (2 (x \cdot x) - 1 \cdot x) + \frac{(C) (x (2 x + 1) (2 x - 1))}{2 x - 1}$

Step 1.1.8.13.1.2

Multiply $x$ by $x$ .

$6 x^{2} - 2 x - 1 = 4 A x^{2} - A + B (2 x^{2} - 1 \cdot x) + \frac{(C) (x (2 x + 1) (2 x - 1))}{2 x - 1}$

Step 1.1.8.13.2

Rewrite $- 1 x$ as $- x$ .

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

Step 1.1.8.14

Apply the distributive property.

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

Step 1.1.8.15

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.16

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.17

Cancel the common factor of $2 x - 1$ .

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Step 1.1.8.17.1

Cancel the common factor.

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

Step 1.1.8.17.2

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

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

Step 1.1.8.18

Apply the distributive property.

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

Step 1.1.8.19

Rewrite using the commutative property of multiplication.

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

Step 1.1.8.20

Multiply $x$ by $1$ .

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

Step 1.1.8.21

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

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Step 1.1.8.21.1

Move $x$ .

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

Step 1.1.8.21.2

Multiply $x$ by $x$ .

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

Step 1.1.8.22

Apply the distributive property.

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

Step 1.1.8.23

Rewrite using the commutative property of multiplication.

$6 x^{2} - 2 x - 1 = 4 A x^{2} - A + 2 B x^{2} - B x + 2 C x^{2} + C x$

Step 1.1.9

Reorder.

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Step 1.1.9.1

Move $A$ .

$6 x^{2} - 2 x - 1 = 4 x^{2} A - A + 2 B x^{2} - B x + 2 C x^{2} + C x$

Step 1.1.9.2

Move $B$ .

$6 x^{2} - 2 x - 1 = 4 x^{2} A - A + 2 x^{2} B - B x + 2 C x^{2} + C x$

Step 1.1.9.3

Move $B$ .

$6 x^{2} - 2 x - 1 = 4 x^{2} A - A + 2 x^{2} B - x B + 2 C x^{2} + C x$

Step 1.1.9.4

Move $- A$ .

$6 x^{2} - 2 x - 1 = 4 x^{2} A + 2 x^{2} B - x B + 2 C x^{2} + C x - A$

Step 1.1.9.5

Move $- x B$ .

$6 x^{2} - 2 x - 1 = 4 x^{2} A + 2 x^{2} B + 2 C x^{2} - x B + C x - A$

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

$6 = 4 A + 2 B + 2 C$

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.

$- 2 = - 1 B + C$

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.

$- 1 = - 1 A$

Step 1.2.4

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

$6 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

$- 1 = - 1 A$

$6 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

$- 1 = - 1 A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $A$ in $- 1 = - 1 A$ .

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Step 1.3.1.1

Rewrite the equation as $- 1 A = - 1$ .

$- 1 A = - 1$

$6 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

Step 1.3.1.2

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

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Step 1.3.1.2.1

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

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

$6 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

Step 1.3.1.2.2

Simplify the left side.

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Step 1.3.1.2.2.1

Dividing two negative values results in a positive value.

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

$6 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

Step 1.3.1.2.2.2

Divide $A$ by $1$ .

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

$6 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

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

$6 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

Step 1.3.1.2.3

Simplify the right side.

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Step 1.3.1.2.3.1

Divide $- 1$ by $- 1$ .

$A = 1$

$6 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

$A = 1$

$6 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

$A = 1$

$6 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

$A = 1$

$6 = 4 A + 2 B + 2 C$

$- 2 = - 1 B + C$

Step 1.3.2

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

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Step 1.3.2.1

Replace all occurrences of $A$ in $6 = 4 A + 2 B + 2 C$ with $1$ .

$6 = 4 (1) + 2 B + 2 C$

$A = 1$

$- 2 = - 1 B + C$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Multiply $4$ by $1$ .

$6 = 4 + 2 B + 2 C$

$A = 1$

$- 2 = - 1 B + C$

$6 = 4 + 2 B + 2 C$

$A = 1$

$- 2 = - 1 B + C$

$6 = 4 + 2 B + 2 C$

$A = 1$

$- 2 = - 1 B + C$

Step 1.3.3

Solve for $B$ in $6 = 4 + 2 B + 2 C$ .

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Step 1.3.3.1

Rewrite the equation as $4 + 2 B + 2 C = 6$ .

$4 + 2 B + 2 C = 6$

$A = 1$

$- 2 = - 1 B + C$

Step 1.3.3.2

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

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Step 1.3.3.2.1

Subtract $4$ from both sides of the equation.

$2 B + 2 C = 6 - 4$

$A = 1$

$- 2 = - 1 B + C$

Step 1.3.3.2.2

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

$2 B = 6 - 4 - 2 C$

$A = 1$

$- 2 = - 1 B + C$

Step 1.3.3.2.3

Subtract $4$ from $6$ .

$2 B = 2 - 2 C$

$A = 1$

$- 2 = - 1 B + C$

$2 B = 2 - 2 C$

$A = 1$

$- 2 = - 1 B + C$

Step 1.3.3.3

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

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Step 1.3.3.3.1

Divide each term in $2 B = 2 - 2 C$ by $2$ .

$\frac{2 B}{2} = \frac{2}{2} + \frac{- 2 C}{2}$

$A = 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.2

Simplify the left side.

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Step 1.3.3.3.2.1

Cancel the common factor of $2$ .

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Step 1.3.3.3.2.1.1

Cancel the common factor.

$\frac{2 B}{2} = \frac{2}{2} + \frac{- 2 C}{2}$

$A = 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{2}{2} + \frac{- 2 C}{2}$

$A = 1$

$- 2 = - 1 B + C$

$B = \frac{2}{2} + \frac{- 2 C}{2}$

$A = 1$

$- 2 = - 1 B + C$

$B = \frac{2}{2} + \frac{- 2 C}{2}$

$A = 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Simplify each term.

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Step 1.3.3.3.3.1.1

Divide $2$ by $2$ .

$B = 1 + \frac{- 2 C}{2}$

$A = 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.3.1.2

Cancel the common factor of $- 2$ and $2$ .

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Step 1.3.3.3.3.1.2.1

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

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

$A = 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.3.1.2.2

Cancel the common factors.

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Step 1.3.3.3.3.1.2.2.1

Factor $2$ out of $2$ .

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

$A = 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.3.1.2.2.2

Cancel the common factor.

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

$A = 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.3.1.2.2.3

Rewrite the expression.

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

$A = 1$

$- 2 = - 1 B + C$

Step 1.3.3.3.3.1.2.2.4

Divide $- C$ by $1$ .

$B = 1 - C$

$A = 1$

$- 2 = - 1 B + C$

$B = 1 - C$

$A = 1$

$- 2 = - 1 B + C$

$B = 1 - C$

$A = 1$

$- 2 = - 1 B + C$

$B = 1 - C$

$A = 1$

$- 2 = - 1 B + C$

$B = 1 - C$

$A = 1$

$- 2 = - 1 B + C$

$B = 1 - C$

$A = 1$

$- 2 = - 1 B + C$

$B = 1 - C$

$A = 1$

$- 2 = - 1 B + C$

Step 1.3.4

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

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Step 1.3.4.1

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

$- 2 = - 1 (1 - C) + C$

$B = 1 - C$

$A = 1$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $- 1 (1 - C) + C$ .

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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.

$- 2 = - 1 \cdot 1 - 1 (- C) + C$

$B = 1 - C$

$A = 1$

Step 1.3.4.2.1.1.2

Multiply $- 1$ by $1$ .

$- 2 = - 1 - 1 (- C) + C$

$B = 1 - C$

$A = 1$

Step 1.3.4.2.1.1.3

Multiply $- 1 (- C)$ .

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Step 1.3.4.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$- 2 = - 1 + 1 C + C$

$B = 1 - C$

$A = 1$

Step 1.3.4.2.1.1.3.2

Multiply $C$ by $1$ .

$- 2 = - 1 + C + C$

$B = 1 - C$

$A = 1$

$- 2 = - 1 + C + C$

$B = 1 - C$

$A = 1$

$- 2 = - 1 + C + C$

$B = 1 - C$

$A = 1$

Step 1.3.4.2.1.2

Add $C$ and $C$ .

$- 2 = - 1 + 2 C$

$B = 1 - C$

$A = 1$

$- 2 = - 1 + 2 C$

$B = 1 - C$

$A = 1$

$- 2 = - 1 + 2 C$

$B = 1 - C$

$A = 1$

$- 2 = - 1 + 2 C$

$B = 1 - C$

$A = 1$

Step 1.3.5

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

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Step 1.3.5.1

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

$- 1 + 2 C = - 2$

$B = 1 - C$

$A = 1$

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 $1$ to both sides of the equation.

$2 C = - 2 + 1$

$B = 1 - C$

$A = 1$

Step 1.3.5.2.2

Add $- 2$ and $1$ .

$2 C = - 1$

$B = 1 - C$

$A = 1$

$2 C = - 1$

$B = 1 - C$

$A = 1$

Step 1.3.5.3

Divide each term in $2 C = - 1$ by $2$ and simplify.

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Step 1.3.5.3.1

Divide each term in $2 C = - 1$ by $2$ .

$\frac{2 C}{2} = \frac{- 1}{2}$

$B = 1 - C$

$A = 1$

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 $2$ .

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Step 1.3.5.3.2.1.1

Cancel the common factor.

$\frac{2 C}{2} = \frac{- 1}{2}$

$B = 1 - C$

$A = 1$

Step 1.3.5.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{- 1}{2}$

$B = 1 - C$

$A = 1$

$C = \frac{- 1}{2}$

$B = 1 - C$

$A = 1$

$C = \frac{- 1}{2}$

$B = 1 - C$

$A = 1$

Step 1.3.5.3.3

Simplify the right side.

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Step 1.3.5.3.3.1

Move the negative in front of the fraction.

$C = - \frac{1}{2}$

$B = 1 - C$

$A = 1$

$C = - \frac{1}{2}$

$B = 1 - C$

$A = 1$

$C = - \frac{1}{2}$

$B = 1 - C$

$A = 1$

$C = - \frac{1}{2}$

$B = 1 - C$

$A = 1$

Step 1.3.6

Replace all occurrences of $C$ with $- \frac{1}{2}$ in each equation.

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Step 1.3.6.1

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

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

$C = - \frac{1}{2}$

$A = 1$

Step 1.3.6.2

Simplify the right side.

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Step 1.3.6.2.1

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

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Step 1.3.6.2.1.1

Multiply $- (- \frac{1}{2})$ .

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Step 1.3.6.2.1.1.1

Multiply $- 1$ by $- 1$ .

$B = 1 + 1 (\frac{1}{2})$

$C = - \frac{1}{2}$

$A = 1$

Step 1.3.6.2.1.1.2

Multiply $\frac{1}{2}$ by $1$ .

$B = 1 + \frac{1}{2}$

$C = - \frac{1}{2}$

$A = 1$

$B = 1 + \frac{1}{2}$

$C = - \frac{1}{2}$

$A = 1$

Step 1.3.6.2.1.2

Write $1$ as a fraction with a common denominator.

$B = \frac{2}{2} + \frac{1}{2}$

$C = - \frac{1}{2}$

$A = 1$

Step 1.3.6.2.1.3

Combine the numerators over the common denominator.

$B = \frac{2 + 1}{2}$

$C = - \frac{1}{2}$

$A = 1$

Step 1.3.6.2.1.4

Add $2$ and $1$ .

$B = \frac{3}{2}$

$C = - \frac{1}{2}$

$A = 1$

$B = \frac{3}{2}$

$C = - \frac{1}{2}$

$A = 1$

$B = \frac{3}{2}$

$C = - \frac{1}{2}$

$A = 1$

$B = \frac{3}{2}$

$C = - \frac{1}{2}$

$A = 1$

Step 1.3.7

List all of the solutions.

$B = \frac{3}{2}, C = - \frac{1}{2}, A = 1$

Step 1.4

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

$\frac{1}{x} + \frac{\frac{3}{2}}{2 x + 1} + \frac{- \frac{1}{2}}{2 x - 1}$

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{x} + \frac{3}{2} \cdot \frac{1}{2 x + 1} + \frac{- \frac{1}{2}}{2 x - 1}$

Step 1.5.2

Multiply $\frac{3}{2}$ by $\frac{1}{2 x + 1}$ .

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

Step 1.5.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.5.4

Multiply $\frac{1}{2 x - 1}$ by $\frac{1}{2}$ .

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

Step 1.5.5

Move $2$ to the left of $2 x - 1$ .

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

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

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

Step 4

Since $\frac{3}{2}$ is constant with respect to $x$ , move $\frac{3}{2}$ out of the integral.

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

Step 5

Let $u_{1} = 2 x + 1$ . Then $d u_{1} = 2 d x$ , so $\frac{1}{2} d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 5.1

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

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Step 5.1.1

Differentiate $2 x + 1$ .

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

Step 5.1.2

By the Sum Rule, the derivative of $2 x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [1]$ .

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

Step 5.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 5.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 5.1.3.2

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

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

Step 5.1.3.3

Multiply $2$ by $1$ .

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

Step 5.1.4

Differentiate using the Constant Rule.

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Step 5.1.4.1

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

$2 + 0$

Step 5.1.4.2

Add $2$ and $0$ .

$2$

Step 5.2

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

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

Step 6

Simplify.

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Step 6.1

Multiply $\frac{1}{u_{1}}$ by $\frac{1}{2}$ .

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

Step 6.2

Move $2$ to the left of $u_{1}$ .

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

Step 7

Since $\frac{1}{2}$ is constant with respect to $u_{1}$ , move $\frac{1}{2}$ out of the integral.

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

Step 8

Simplify.

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Step 8.1

Multiply $\frac{1}{2}$ by $\frac{3}{2}$ .

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

Step 8.2

Multiply $2$ by $2$ .

$\ln (| x |) + C + \frac{3}{4} \int \frac{1}{u_{1}} d u_{1} + \int - \frac{1}{2 (2 x - 1)} d x$

Step 9

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

$\ln (| x |) + C + \frac{3}{4} (\ln (| u_{1} |) + C) + \int - \frac{1}{2 (2 x - 1)} d x$

Step 10

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

$\ln (| x |) + C + \frac{3}{4} (\ln (| u_{1} |) + C) - \int \frac{1}{2 (2 x - 1)} d x$

Step 11

Since $\frac{1}{2}$ is constant with respect to $x$ , move $\frac{1}{2}$ out of the integral.

$\ln (| x |) + C + \frac{3}{4} (\ln (| u_{1} |) + C) - (\frac{1}{2} \int \frac{1}{2 x - 1} d x)$

Step 12

Let $u_{2} = 2 x - 1$ . Then $d u_{2} = 2 d x$ , so $\frac{1}{2} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 12.1

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

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Step 12.1.1

Differentiate $2 x - 1$ .

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

Step 12.1.2

By the Sum Rule, the derivative of $2 x - 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$ .

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

Step 12.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 12.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 12.1.3.2

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

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

Step 12.1.3.3

Multiply $2$ by $1$ .

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

Step 12.1.4

Differentiate using the Constant Rule.

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Step 12.1.4.1

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

$2 + 0$

Step 12.1.4.2

Add $2$ and $0$ .

$2$

Step 12.2

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

$\ln (| x |) + C + \frac{3}{4} (\ln (| u_{1} |) + C) - \frac{1}{2} \int \frac{1}{u_{2}} \cdot \frac{1}{2} d u_{2}$

Step 13

Simplify.

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Step 13.1

Multiply $\frac{1}{u_{2}}$ by $\frac{1}{2}$ .

$\ln (| x |) + C + \frac{3}{4} (\ln (| u_{1} |) + C) - \frac{1}{2} \int \frac{1}{u_{2} \cdot 2} d u_{2}$

Step 13.2

Move $2$ to the left of $u_{2}$ .

$\ln (| x |) + C + \frac{3}{4} (\ln (| u_{1} |) + C) - \frac{1}{2} \int \frac{1}{2 u_{2}} d u_{2}$

Step 14

Since $\frac{1}{2}$ is constant with respect to $u_{2}$ , move $\frac{1}{2}$ out of the integral.

$\ln (| x |) + C + \frac{3}{4} (\ln (| u_{1} |) + C) - \frac{1}{2} (\frac{1}{2} \int \frac{1}{u_{2}} d u_{2})$

Step 15

Simplify.

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Step 15.1

Multiply $\frac{1}{2}$ by $\frac{1}{2}$ .

$\ln (| x |) + C + \frac{3}{4} (\ln (| u_{1} |) + C) - \frac{1}{2 \cdot 2} \int \frac{1}{u_{2}} d u_{2}$

Step 15.2

Multiply $2$ by $2$ .

$\ln (| x |) + C + \frac{3}{4} (\ln (| u_{1} |) + C) - \frac{1}{4} \int \frac{1}{u_{2}} d u_{2}$

Step 16

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

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

Step 17

Simplify.

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

Step 18

Substitute back in for each integration substitution variable.

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Step 18.1

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

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

Step 18.2

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

$\ln (| x |) + \frac{3}{4} \ln (| 2 x + 1 |) - \frac{1}{4} \ln (| 2 x - 1 |) + C$