Calculus Examples

$\int \frac{2 x + 1}{(3 x - 1) (2 x + 5)} d x$

Step 1

Write the fraction using partial fraction decomposition.

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Step 1.1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1.1

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

$\frac{A}{3 x - 1}$

Step 1.1.2

$\frac{A}{3 x - 1} + \frac{B}{2 x + 5}$

Step 1.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(3 x - 1) (2 x + 5)$ .

$\frac{(2 x + 1) (3 x - 1) (2 x + 5)}{(3 x - 1) (2 x + 5)} = \frac{(A) (3 x - 1) (2 x + 5)}{3 x - 1} + \frac{(B) (3 x - 1) (2 x + 5)}{2 x + 5}$

Step 1.1.4

Cancel the common factor of $3 x - 1$ .

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Step 1.1.4.1

Cancel the common factor.

$\frac{(2 x + 1) (3 x - 1) (2 x + 5)}{(3 x - 1) (2 x + 5)} = \frac{(A) (3 x - 1) (2 x + 5)}{3 x - 1} + \frac{(B) (3 x - 1) (2 x + 5)}{2 x + 5}$

Step 1.1.4.2

Rewrite the expression.

$\frac{(2 x + 1) (2 x + 5)}{2 x + 5} = \frac{(A) (3 x - 1) (2 x + 5)}{3 x - 1} + \frac{(B) (3 x - 1) (2 x + 5)}{2 x + 5}$

Step 1.1.5

Cancel the common factor of $2 x + 5$ .

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Step 1.1.5.1

Cancel the common factor.

$\frac{(2 x + 1) (2 x + 5)}{2 x + 5} = \frac{(A) (3 x - 1) (2 x + 5)}{3 x - 1} + \frac{(B) (3 x - 1) (2 x + 5)}{2 x + 5}$

Step 1.1.5.2

Divide $2 x + 1$ by $1$ .

$2 x + 1 = \frac{(A) (3 x - 1) (2 x + 5)}{3 x - 1} + \frac{(B) (3 x - 1) (2 x + 5)}{2 x + 5}$

Step 1.1.6

Simplify each term.

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Step 1.1.6.1

Cancel the common factor of $3 x - 1$ .

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Step 1.1.6.1.1

Cancel the common factor.

$2 x + 1 = \frac{A (3 x - 1) (2 x + 5)}{3 x - 1} + \frac{(B) (3 x - 1) (2 x + 5)}{2 x + 5}$

Step 1.1.6.1.2

Divide $(A) (2 x + 5)$ by $1$ .

$2 x + 1 = (A) (2 x + 5) + \frac{(B) (3 x - 1) (2 x + 5)}{2 x + 5}$

Step 1.1.6.2

Apply the distributive property.

$2 x + 1 = A (2 x) + A \cdot 5 + \frac{(B) (3 x - 1) (2 x + 5)}{2 x + 5}$

Step 1.1.6.3

Rewrite using the commutative property of multiplication.

$2 x + 1 = 2 A x + A \cdot 5 + \frac{(B) (3 x - 1) (2 x + 5)}{2 x + 5}$

Step 1.1.6.4

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

$2 x + 1 = 2 A x + 5 \cdot A + \frac{(B) (3 x - 1) (2 x + 5)}{2 x + 5}$

Step 1.1.6.5

Cancel the common factor of $2 x + 5$ .

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Step 1.1.6.5.1

Cancel the common factor.

$2 x + 1 = 2 A x + 5 A + \frac{(B) (3 x - 1) (2 x + 5)}{2 x + 5}$

Step 1.1.6.5.2

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

$2 x + 1 = 2 A x + 5 A + (B) (3 x - 1)$

Step 1.1.6.6

Apply the distributive property.

$2 x + 1 = 2 A x + 5 A + B (3 x) + B \cdot - 1$

Step 1.1.6.7

Rewrite using the commutative property of multiplication.

$2 x + 1 = 2 A x + 5 A + 3 B x + B \cdot - 1$

Step 1.1.6.8

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

$2 x + 1 = 2 A x + 5 A + 3 B x - 1 \cdot B$

Step 1.1.6.9

Rewrite $- 1 B$ as $- B$ .

$2 x + 1 = 2 A x + 5 A + 3 B x - B$

Step 1.1.7

Reorder.

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Step 1.1.7.1

Move $A$ .

$2 x + 1 = 2 x A + 5 A + 3 B x - B$

Step 1.1.7.2

Move $B$ .

$2 x + 1 = 2 x A + 5 A + 3 x B - B$

Step 1.1.7.3

Move $5 A$ .

$2 x + 1 = 2 x A + 3 x B + 5 A - B$

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

$2 = 2 A + 3 B$

Step 1.2.2

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.

$1 = 5 A - 1 B$

Step 1.2.3

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

$2 = 2 A + 3 B$

$1 = 5 A - 1 B$

$2 = 2 A + 3 B$

$1 = 5 A - 1 B$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $A$ in $2 = 2 A + 3 B$ .

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Step 1.3.1.1

Rewrite the equation as $2 A + 3 B = 2$ .

$2 A + 3 B = 2$

$1 = 5 A - 1 B$

Step 1.3.1.2

Subtract $3 B$ from both sides of the equation.

$2 A = 2 - 3 B$

$1 = 5 A - 1 B$

Step 1.3.1.3

Divide each term in $2 A = 2 - 3 B$ by $2$ and simplify.

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Step 1.3.1.3.1

Divide each term in $2 A = 2 - 3 B$ by $2$ .

$\frac{2 A}{2} = \frac{2}{2} + \frac{- 3 B}{2}$

$1 = 5 A - 1 B$

Step 1.3.1.3.2

Simplify the left side.

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Step 1.3.1.3.2.1

Cancel the common factor of $2$ .

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Step 1.3.1.3.2.1.1

Cancel the common factor.

$\frac{2 A}{2} = \frac{2}{2} + \frac{- 3 B}{2}$

$1 = 5 A - 1 B$

Step 1.3.1.3.2.1.2

Divide $A$ by $1$ .

$A = \frac{2}{2} + \frac{- 3 B}{2}$

$1 = 5 A - 1 B$

$A = \frac{2}{2} + \frac{- 3 B}{2}$

$1 = 5 A - 1 B$

$A = \frac{2}{2} + \frac{- 3 B}{2}$

$1 = 5 A - 1 B$

Step 1.3.1.3.3

Simplify the right side.

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Step 1.3.1.3.3.1

Simplify each term.

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Step 1.3.1.3.3.1.1

Divide $2$ by $2$ .

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

$1 = 5 A - 1 B$

Step 1.3.1.3.3.1.2

Move the negative in front of the fraction.

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

$1 = 5 A - 1 B$

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

$1 = 5 A - 1 B$

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

$1 = 5 A - 1 B$

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

$1 = 5 A - 1 B$

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

$1 = 5 A - 1 B$

Step 1.3.2

Replace all occurrences of $A$ with $1 - \frac{3 B}{2}$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $A$ in $1 = 5 A - 1 B$ with $1 - \frac{3 B}{2}$ .

$1 = 5 (1 - \frac{3 B}{2}) - 1 B$

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

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Simplify $5 (1 - \frac{3 B}{2}) - 1 B$ .

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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.

$1 = 5 \cdot 1 + 5 (- \frac{3 B}{2}) - 1 B$

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

Step 1.3.2.2.1.1.2

Multiply $5$ by $1$ .

$1 = 5 + 5 (- \frac{3 B}{2}) - 1 B$

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

Step 1.3.2.2.1.1.3

Multiply $5 (- \frac{3 B}{2})$ .

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Step 1.3.2.2.1.1.3.1

Multiply $- 1$ by $5$ .

$1 = 5 - 5 \frac{3 B}{2} - 1 B$

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

Step 1.3.2.2.1.1.3.2

Combine $- 5$ and $\frac{3 B}{2}$ .

$1 = 5 + \frac{- 5 (3 B)}{2} - 1 B$

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

Step 1.3.2.2.1.1.3.3

Multiply $3$ by $- 5$ .

$1 = 5 + \frac{- 15 B}{2} - 1 B$

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

$1 = 5 + \frac{- 15 B}{2} - 1 B$

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

Step 1.3.2.2.1.1.4

Move the negative in front of the fraction.

$1 = 5 - \frac{15 B}{2} - 1 B$

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

Step 1.3.2.2.1.1.5

Rewrite $- 1 B$ as $- B$ .

$1 = 5 - \frac{15 B}{2} - B$

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

$1 = 5 - \frac{15 B}{2} - B$

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

Step 1.3.2.2.1.2

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

$1 = 5 - \frac{15 B}{2} - B \cdot \frac{2}{2}$

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

Step 1.3.2.2.1.3

Simplify terms.

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Step 1.3.2.2.1.3.1

Combine $- B$ and $\frac{2}{2}$ .

$1 = 5 - \frac{15 B}{2} + \frac{- B \cdot 2}{2}$

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

Step 1.3.2.2.1.3.2

Combine the numerators over the common denominator.

$1 = 5 + \frac{- 15 B - B \cdot 2}{2}$

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

$1 = 5 + \frac{- 15 B - B \cdot 2}{2}$

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

Step 1.3.2.2.1.4

Simplify each term.

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Step 1.3.2.2.1.4.1

Simplify the numerator.

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Step 1.3.2.2.1.4.1.1

Factor $B$ out of $- 15 B - B \cdot 2$ .

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Step 1.3.2.2.1.4.1.1.1

Factor $B$ out of $- 15 B$ .

$1 = 5 + \frac{B \cdot - 15 - B \cdot 2}{2}$

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

Step 1.3.2.2.1.4.1.1.2

Factor $B$ out of $- B \cdot 2$ .

$1 = 5 + \frac{B \cdot - 15 + B (- 1 \cdot 2)}{2}$

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

Step 1.3.2.2.1.4.1.1.3

Factor $B$ out of $B \cdot - 15 + B (- 1 \cdot 2)$ .

$1 = 5 + \frac{B (- 15 - 1 \cdot 2)}{2}$

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

$1 = 5 + \frac{B (- 15 - 1 \cdot 2)}{2}$

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

Step 1.3.2.2.1.4.1.2

Multiply $- 1$ by $2$ .

$1 = 5 + \frac{B (- 15 - 2)}{2}$

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

Step 1.3.2.2.1.4.1.3

Subtract $2$ from $- 15$ .

$1 = 5 + \frac{B \cdot - 17}{2}$

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

$1 = 5 + \frac{B \cdot - 17}{2}$

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

Step 1.3.2.2.1.4.2

Move $- 17$ to the left of $B$ .

$1 = 5 + \frac{- 17 \cdot B}{2}$

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

Step 1.3.2.2.1.4.3

Move the negative in front of the fraction.

$1 = 5 - \frac{17 B}{2}$

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

$1 = 5 - \frac{17 B}{2}$

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

$1 = 5 - \frac{17 B}{2}$

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

$1 = 5 - \frac{17 B}{2}$

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

$1 = 5 - \frac{17 B}{2}$

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

Step 1.3.3

Solve for $B$ in $1 = 5 - \frac{17 B}{2}$ .

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Step 1.3.3.1

Rewrite the equation as $5 - \frac{17 B}{2} = 1$ .

$5 - \frac{17 B}{2} = 1$

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

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 $5$ from both sides of the equation.

$- \frac{17 B}{2} = 1 - 5$

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

Step 1.3.3.2.2

Subtract $5$ from $1$ .

$- \frac{17 B}{2} = - 4$

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

$- \frac{17 B}{2} = - 4$

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

Step 1.3.3.3

Multiply both sides of the equation by $- \frac{2}{17}$ .

$- \frac{2}{17} \cdot (- \frac{17 B}{2}) = - \frac{2}{17} \cdot - 4$

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

Step 1.3.3.4

Simplify both sides of the equation.

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Step 1.3.3.4.1

Simplify the left side.

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Step 1.3.3.4.1.1

Simplify $- \frac{2}{17} (- \frac{17 B}{2})$ .

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Step 1.3.3.4.1.1.1

Cancel the common factor of $2$ .

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Step 1.3.3.4.1.1.1.1

Move the leading negative in $- \frac{2}{17}$ into the numerator.

$\frac{- 2}{17} \cdot (- \frac{17 B}{2}) = - \frac{2}{17} \cdot - 4$

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

Step 1.3.3.4.1.1.1.2

Move the leading negative in $- \frac{17 B}{2}$ into the numerator.

$\frac{- 2}{17} \cdot \frac{- 17 B}{2} = - \frac{2}{17} \cdot - 4$

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

Step 1.3.3.4.1.1.1.3

Factor $2$ out of $- 2$ .

$\frac{2 (- 1)}{17} \cdot \frac{- 17 B}{2} = - \frac{2}{17} \cdot - 4$

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

Step 1.3.3.4.1.1.1.4

Cancel the common factor.

$\frac{2 \cdot - 1}{17} \cdot \frac{- 17 B}{2} = - \frac{2}{17} \cdot - 4$

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

Step 1.3.3.4.1.1.1.5

Rewrite the expression.

$\frac{- 1}{17} \cdot (- 17 B) = - \frac{2}{17} \cdot - 4$

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

$\frac{- 1}{17} \cdot (- 17 B) = - \frac{2}{17} \cdot - 4$

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

Step 1.3.3.4.1.1.2

Cancel the common factor of $17$ .

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Step 1.3.3.4.1.1.2.1

Factor $17$ out of $- 17 B$ .

$\frac{- 1}{17} \cdot (17 (- B)) = - \frac{2}{17} \cdot - 4$

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

Step 1.3.3.4.1.1.2.2

Cancel the common factor.

$\frac{- 1}{17} \cdot (17 (- B)) = - \frac{2}{17} \cdot - 4$

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

Step 1.3.3.4.1.1.2.3

Rewrite the expression.

$B = - \frac{2}{17} \cdot - 4$

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

$B = - \frac{2}{17} \cdot - 4$

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

Step 1.3.3.4.1.1.3

Multiply.

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Step 1.3.3.4.1.1.3.1

Multiply $- 1$ by $- 1$ .

$1 B = - \frac{2}{17} \cdot - 4$

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

Step 1.3.3.4.1.1.3.2

Multiply $B$ by $1$ .

$B = - \frac{2}{17} \cdot - 4$

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

$B = - \frac{2}{17} \cdot - 4$

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

$B = - \frac{2}{17} \cdot - 4$

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

$B = - \frac{2}{17} \cdot - 4$

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

Step 1.3.3.4.2

Simplify the right side.

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Step 1.3.3.4.2.1

Multiply $- \frac{2}{17} \cdot - 4$ .

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Step 1.3.3.4.2.1.1

Multiply $- 4$ by $- 1$ .

$B = 4 (\frac{2}{17})$

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

Step 1.3.3.4.2.1.2

Combine $4$ and $\frac{2}{17}$ .

$B = \frac{4 \cdot 2}{17}$

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

Step 1.3.3.4.2.1.3

Multiply $4$ by $2$ .

$B = \frac{8}{17}$

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

$B = \frac{8}{17}$

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

$B = \frac{8}{17}$

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

$B = \frac{8}{17}$

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

$B = \frac{8}{17}$

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

Step 1.3.4

Replace all occurrences of $B$ with $\frac{8}{17}$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $B$ in $A = 1 - \frac{3 B}{2}$ with $\frac{8}{17}$ .

$A = 1 - \frac{3 (\frac{8}{17})}{2}$

$B = \frac{8}{17}$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $1 - \frac{3 (\frac{8}{17})}{2}$ .

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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

Combine $3$ and $\frac{8}{17}$ .

$A = 1 - \frac{\frac{3 \cdot 8}{17}}{2}$

$B = \frac{8}{17}$

Step 1.3.4.2.1.1.2

Multiply $3$ by $8$ .

$A = 1 - \frac{\frac{24}{17}}{2}$

$B = \frac{8}{17}$

Step 1.3.4.2.1.1.3

Multiply the numerator by the reciprocal of the denominator.

$A = 1 - (\frac{24}{17} \cdot \frac{1}{2})$

$B = \frac{8}{17}$

Step 1.3.4.2.1.1.4

Cancel the common factor of $2$ .

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Step 1.3.4.2.1.1.4.1

Factor $2$ out of $24$ .

$A = 1 - (\frac{2 (12)}{17} \cdot \frac{1}{2})$

$B = \frac{8}{17}$

Step 1.3.4.2.1.1.4.2

Cancel the common factor.

$A = 1 - (\frac{2 \cdot 12}{17} \cdot \frac{1}{2})$

$B = \frac{8}{17}$

Step 1.3.4.2.1.1.4.3

Rewrite the expression.

$A = 1 - \frac{12}{17}$

$B = \frac{8}{17}$

$A = 1 - \frac{12}{17}$

$B = \frac{8}{17}$

$A = 1 - \frac{12}{17}$

$B = \frac{8}{17}$

Step 1.3.4.2.1.2

Simplify the expression.

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Step 1.3.4.2.1.2.1

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

$A = \frac{17}{17} - \frac{12}{17}$

$B = \frac{8}{17}$

Step 1.3.4.2.1.2.2

Combine the numerators over the common denominator.

$A = \frac{17 - 12}{17}$

$B = \frac{8}{17}$

Step 1.3.4.2.1.2.3

Subtract $12$ from $17$ .

$A = \frac{5}{17}$

$B = \frac{8}{17}$

$A = \frac{5}{17}$

$B = \frac{8}{17}$

$A = \frac{5}{17}$

$B = \frac{8}{17}$

$A = \frac{5}{17}$

$B = \frac{8}{17}$

$A = \frac{5}{17}$

$B = \frac{8}{17}$

Step 1.3.5

List all of the solutions.

$A = \frac{5}{17}, B = \frac{8}{17}$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{3 x - 1} + \frac{B}{2 x + 5}$ with the values found for $A$ and $B$ .

$\frac{\frac{5}{17}}{3 x - 1} + \frac{\frac{8}{17}}{2 x + 5}$

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{5}{17} \cdot \frac{1}{3 x - 1} + \frac{\frac{8}{17}}{2 x + 5}$

Step 1.5.2

Multiply $\frac{5}{17}$ by $\frac{1}{3 x - 1}$ .

$\frac{5}{17 (3 x - 1)} + \frac{\frac{8}{17}}{2 x + 5}$

Step 1.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{5}{17 (3 x - 1)} + \frac{8}{17} \cdot \frac{1}{2 x + 5}$

Step 1.5.4

Multiply $\frac{8}{17}$ by $\frac{1}{2 x + 5}$ .

$\int \frac{5}{17 (3 x - 1)} + \frac{8}{17 (2 x + 5)} d x$

Step 2

Split the single integral into multiple integrals.

$\int \frac{5}{17 (3 x - 1)} d x + \int \frac{8}{17 (2 x + 5)} d x$

Step 3

Since $\frac{5}{17}$ is constant with respect to $x$ , move $\frac{5}{17}$ out of the integral.

$\frac{5}{17} \int \frac{1}{3 x - 1} d x + \int \frac{8}{17 (2 x + 5)} d x$

Step 4

Let $u_{1} = 3 x - 1$ . Then $d u_{1} = 3 d x$ , so $\frac{1}{3} d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 4.1

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

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Step 4.1.1

Differentiate $3 x - 1$ .

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

Step 4.1.2

By the Sum Rule, the derivative of $3 x - 1$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [- 1]$ .

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

Step 4.1.3

Evaluate $\frac{d}{d x} [3 x]$ .

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Step 4.1.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

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

Step 4.1.3.2

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

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

Step 4.1.3.3

Multiply $3$ by $1$ .

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

Step 4.1.4

Differentiate using the Constant Rule.

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Step 4.1.4.1

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

$3 + 0$

Step 4.1.4.2

Add $3$ and $0$ .

$3$

Step 4.2

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

$\frac{5}{17} \int \frac{1}{u_{1}} \cdot \frac{1}{3} d u_{1} + \int \frac{8}{17 (2 x + 5)} d x$

Step 5

Simplify.

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Step 5.1

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

$\frac{5}{17} \int \frac{1}{u_{1} \cdot 3} d u_{1} + \int \frac{8}{17 (2 x + 5)} d x$

Step 5.2

Move $3$ to the left of $u_{1}$ .

$\frac{5}{17} \int \frac{1}{3 u_{1}} d u_{1} + \int \frac{8}{17 (2 x + 5)} d x$

Step 6

Since $\frac{1}{3}$ is constant with respect to $u_{1}$ , move $\frac{1}{3}$ out of the integral.

$\frac{5}{17} (\frac{1}{3} \int \frac{1}{u_{1}} d u_{1}) + \int \frac{8}{17 (2 x + 5)} d x$

Step 7

Simplify.

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Step 7.1

Multiply $\frac{1}{3}$ by $\frac{5}{17}$ .

$\frac{5}{3 \cdot 17} \int \frac{1}{u_{1}} d u_{1} + \int \frac{8}{17 (2 x + 5)} d x$

Step 7.2

Multiply $3$ by $17$ .

$\frac{5}{51} \int \frac{1}{u_{1}} d u_{1} + \int \frac{8}{17 (2 x + 5)} d x$

Step 8

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

$\frac{5}{51} (\ln (| u_{1} |) + C) + \int \frac{8}{17 (2 x + 5)} d x$

Step 9

Since $\frac{8}{17}$ is constant with respect to $x$ , move $\frac{8}{17}$ out of the integral.

$\frac{5}{51} (\ln (| u_{1} |) + C) + \frac{8}{17} \int \frac{1}{2 x + 5} d x$

Step 10

Let $u_{2} = 2 x + 5$ . 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 10.1

Let $u_{2} = 2 x + 5$ . Find $\frac{d u_{2}}{d x}$ .

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Step 10.1.1

Differentiate $2 x + 5$ .

$\frac{d}{d x} [2 x + 5]$

Step 10.1.2

By the Sum Rule, the derivative of $2 x + 5$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [5]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [5]$

Step 10.1.3

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

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Step 10.1.3.1

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

$2 \frac{d}{d x} [x] + \frac{d}{d x} [5]$

Step 10.1.3.2

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

$2 \cdot 1 + \frac{d}{d x} [5]$

Step 10.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [5]$

Step 10.1.4

Differentiate using the Constant Rule.

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Step 10.1.4.1

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

$2 + 0$

Step 10.1.4.2

Add $2$ and $0$ .

$2$

Step 10.2

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

$\frac{5}{51} (\ln (| u_{1} |) + C) + \frac{8}{17} \int \frac{1}{u_{2}} \cdot \frac{1}{2} d u_{2}$

Step 11

Simplify.

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Step 11.1

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

$\frac{5}{51} (\ln (| u_{1} |) + C) + \frac{8}{17} \int \frac{1}{u_{2} \cdot 2} d u_{2}$

Step 11.2

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

$\frac{5}{51} (\ln (| u_{1} |) + C) + \frac{8}{17} \int \frac{1}{2 u_{2}} d u_{2}$

Step 12

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

$\frac{5}{51} (\ln (| u_{1} |) + C) + \frac{8}{17} (\frac{1}{2} \int \frac{1}{u_{2}} d u_{2})$

Step 13

Simplify.

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Step 13.1

Multiply $\frac{1}{2}$ by $\frac{8}{17}$ .

$\frac{5}{51} (\ln (| u_{1} |) + C) + \frac{8}{2 \cdot 17} \int \frac{1}{u_{2}} d u_{2}$

Step 13.2

Multiply $2$ by $17$ .

$\frac{5}{51} (\ln (| u_{1} |) + C) + \frac{8}{34} \int \frac{1}{u_{2}} d u_{2}$

Step 13.3

Cancel the common factor of $8$ and $34$ .

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Step 13.3.1

Factor $2$ out of $8$ .

$\frac{5}{51} (\ln (| u_{1} |) + C) + \frac{2 (4)}{34} \int \frac{1}{u_{2}} d u_{2}$

Step 13.3.2

Cancel the common factors.

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Step 13.3.2.1

Factor $2$ out of $34$ .

$\frac{5}{51} (\ln (| u_{1} |) + C) + \frac{2 \cdot 4}{2 \cdot 17} \int \frac{1}{u_{2}} d u_{2}$

Step 13.3.2.2

Cancel the common factor.

$\frac{5}{51} (\ln (| u_{1} |) + C) + \frac{2 \cdot 4}{2 \cdot 17} \int \frac{1}{u_{2}} d u_{2}$

Step 13.3.2.3

Rewrite the expression.

$\frac{5}{51} (\ln (| u_{1} |) + C) + \frac{4}{17} \int \frac{1}{u_{2}} d u_{2}$

Step 14

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

$\frac{5}{51} (\ln (| u_{1} |) + C) + \frac{4}{17} (\ln (| u_{2} |) + C)$

Step 15

Simplify.

$\frac{5}{51} \ln (| u_{1} |) + \frac{4}{17} \ln (| u_{2} |) + C$

Step 16

Substitute back in for each integration substitution variable.

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Step 16.1

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

$\frac{5}{51} \ln (| 3 x - 1 |) + \frac{4}{17} \ln (| u_{2} |) + C$

Step 16.2

Replace all occurrences of $u_{2}$ with $2 x + 5$ .

$\frac{5}{51} \ln (| 3 x - 1 |) + \frac{4}{17} \ln (| 2 x + 5 |) + C$