Calculus Examples

$\int \frac{7 x}{(2 x - 3) (x + 2)} d x$

Step 1

Since $7$ is constant with respect to $x$ , move $7$ out of the integral.

$7 \int \frac{x}{(2 x - 3) (x + 2)} d x$

Step 2

Write the fraction using partial fraction decomposition.

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Step 2.1

Decompose the fraction and multiply through by the common denominator.

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Step 2.1.1

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 $A$ .

$\frac{A}{2 x - 3}$

Step 2.1.2

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

Step 2.1.3

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

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

Step 2.1.4

Cancel the common factor of $2 x - 3$ .

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Step 2.1.4.1

Cancel the common factor.

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

Step 2.1.4.2

Rewrite the expression.

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

Step 2.1.5

Cancel the common factor of $x + 2$ .

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Step 2.1.5.1

Cancel the common factor.

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

Step 2.1.5.2

Divide $x$ by $1$ .

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

Step 2.1.6

Simplify each term.

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Step 2.1.6.1

Cancel the common factor of $2 x - 3$ .

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Step 2.1.6.1.1

Cancel the common factor.

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

Step 2.1.6.1.2

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

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

Step 2.1.6.2

Apply the distributive property.

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

Step 2.1.6.3

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

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

Step 2.1.6.4

Cancel the common factor of $x + 2$ .

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Step 2.1.6.4.1

Cancel the common factor.

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

Step 2.1.6.4.2

Divide $(B) (2 x - 3)$ by $1$ .

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

Step 2.1.6.5

Apply the distributive property.

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

Step 2.1.6.6

Rewrite using the commutative property of multiplication.

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

Step 2.1.6.7

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

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

Step 2.1.7

Simplify the expression.

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Step 2.1.7.1

Move $B$ .

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

Step 2.1.7.2

Move $2 A$ .

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

Step 2.2

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

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Step 2.2.1

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.

$1 = A + 2 B$

Step 2.2.2

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.

$0 = 2 A - 3 B$

Step 2.2.3

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

$1 = A + 2 B$

$0 = 2 A - 3 B$

$1 = A + 2 B$

$0 = 2 A - 3 B$

Step 2.3

Solve the system of equations.

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Step 2.3.1

Solve for $A$ in $1 = A + 2 B$ .

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Step 2.3.1.1

Rewrite the equation as $A + 2 B = 1$ .

$A + 2 B = 1$

$0 = 2 A - 3 B$

Step 2.3.1.2

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

$A = 1 - 2 B$

$0 = 2 A - 3 B$

$A = 1 - 2 B$

$0 = 2 A - 3 B$

Step 2.3.2

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

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Step 2.3.2.1

Replace all occurrences of $A$ in $0 = 2 A - 3 B$ with $1 - 2 B$ .

$0 = 2 (1 - 2 B) - 3 B$

$A = 1 - 2 B$

Step 2.3.2.2

Simplify the right side.

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Step 2.3.2.2.1

Simplify $2 (1 - 2 B) - 3 B$ .

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Step 2.3.2.2.1.1

Simplify each term.

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Step 2.3.2.2.1.1.1

Apply the distributive property.

$0 = 2 \cdot 1 + 2 (- 2 B) - 3 B$

$A = 1 - 2 B$

Step 2.3.2.2.1.1.2

Multiply $2$ by $1$ .

$0 = 2 + 2 (- 2 B) - 3 B$

$A = 1 - 2 B$

Step 2.3.2.2.1.1.3

Multiply $- 2$ by $2$ .

$0 = 2 - 4 B - 3 B$

$A = 1 - 2 B$

$0 = 2 - 4 B - 3 B$

$A = 1 - 2 B$

Step 2.3.2.2.1.2

Subtract $3 B$ from $- 4 B$ .

$0 = 2 - 7 B$

$A = 1 - 2 B$

$0 = 2 - 7 B$

$A = 1 - 2 B$

$0 = 2 - 7 B$

$A = 1 - 2 B$

$0 = 2 - 7 B$

$A = 1 - 2 B$

Step 2.3.3

Solve for $B$ in $0 = 2 - 7 B$ .

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Step 2.3.3.1

Rewrite the equation as $2 - 7 B = 0$ .

$2 - 7 B = 0$

$A = 1 - 2 B$

Step 2.3.3.2

Subtract $2$ from both sides of the equation.

$- 7 B = - 2$

$A = 1 - 2 B$

Step 2.3.3.3

Divide each term in $- 7 B = - 2$ by $- 7$ and simplify.

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Step 2.3.3.3.1

Divide each term in $- 7 B = - 2$ by $- 7$ .

$\frac{- 7 B}{- 7} = \frac{- 2}{- 7}$

$A = 1 - 2 B$

Step 2.3.3.3.2

Simplify the left side.

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Step 2.3.3.3.2.1

Cancel the common factor of $- 7$ .

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Step 2.3.3.3.2.1.1

Cancel the common factor.

$\frac{- 7 B}{- 7} = \frac{- 2}{- 7}$

$A = 1 - 2 B$

Step 2.3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{- 2}{- 7}$

$A = 1 - 2 B$

$B = \frac{- 2}{- 7}$

$A = 1 - 2 B$

$B = \frac{- 2}{- 7}$

$A = 1 - 2 B$

Step 2.3.3.3.3

Simplify the right side.

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Step 2.3.3.3.3.1

Dividing two negative values results in a positive value.

$B = \frac{2}{7}$

$A = 1 - 2 B$

$B = \frac{2}{7}$

$A = 1 - 2 B$

$B = \frac{2}{7}$

$A = 1 - 2 B$

$B = \frac{2}{7}$

$A = 1 - 2 B$

Step 2.3.4

Replace all occurrences of $B$ with $\frac{2}{7}$ in each equation.

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Step 2.3.4.1

Replace all occurrences of $B$ in $A = 1 - 2 B$ with $\frac{2}{7}$ .

$A = 1 - 2 (\frac{2}{7})$

$B = \frac{2}{7}$

Step 2.3.4.2

Simplify the right side.

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Step 2.3.4.2.1

Simplify $1 - 2 (\frac{2}{7})$ .

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Step 2.3.4.2.1.1

Simplify each term.

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Step 2.3.4.2.1.1.1

Multiply $- 2 (\frac{2}{7})$ .

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Step 2.3.4.2.1.1.1.1

Combine $- 2$ and $\frac{2}{7}$ .

$A = 1 + \frac{- 2 \cdot 2}{7}$

$B = \frac{2}{7}$

Step 2.3.4.2.1.1.1.2

Multiply $- 2$ by $2$ .

$A = 1 + \frac{- 4}{7}$

$B = \frac{2}{7}$

$A = 1 + \frac{- 4}{7}$

$B = \frac{2}{7}$

Step 2.3.4.2.1.1.2

Move the negative in front of the fraction.

$A = 1 - \frac{4}{7}$

$B = \frac{2}{7}$

$A = 1 - \frac{4}{7}$

$B = \frac{2}{7}$

Step 2.3.4.2.1.2

Simplify the expression.

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Step 2.3.4.2.1.2.1

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

$A = \frac{7}{7} - \frac{4}{7}$

$B = \frac{2}{7}$

Step 2.3.4.2.1.2.2

Combine the numerators over the common denominator.

$A = \frac{7 - 4}{7}$

$B = \frac{2}{7}$

Step 2.3.4.2.1.2.3

Subtract $4$ from $7$ .

$A = \frac{3}{7}$

$B = \frac{2}{7}$

$A = \frac{3}{7}$

$B = \frac{2}{7}$

$A = \frac{3}{7}$

$B = \frac{2}{7}$

$A = \frac{3}{7}$

$B = \frac{2}{7}$

$A = \frac{3}{7}$

$B = \frac{2}{7}$

Step 2.3.5

List all of the solutions.

$A = \frac{3}{7}, B = \frac{2}{7}$

Step 2.4

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

$\frac{\frac{3}{7}}{2 x - 3} + \frac{\frac{2}{7}}{x + 2}$

Step 2.5

Simplify.

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Step 2.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{3}{7} \cdot \frac{1}{2 x - 3} + \frac{\frac{2}{7}}{x + 2}$

Step 2.5.2

Multiply $\frac{3}{7}$ by $\frac{1}{2 x - 3}$ .

$\frac{3}{7 (2 x - 3)} + \frac{\frac{2}{7}}{x + 2}$

Step 2.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{3}{7 (2 x - 3)} + \frac{2}{7} \cdot \frac{1}{x + 2}$

Step 2.5.4

Multiply $\frac{2}{7}$ by $\frac{1}{x + 2}$ .

$7 \int \frac{3}{7 (2 x - 3)} + \frac{2}{7 (x + 2)} d x$

Step 3

Split the single integral into multiple integrals.

$7 (\int \frac{3}{7 (2 x - 3)} d x + \int \frac{2}{7 (x + 2)} d x)$

Step 4

Since $\frac{3}{7}$ is constant with respect to $x$ , move $\frac{3}{7}$ out of the integral.

$7 (\frac{3}{7} \int \frac{1}{2 x - 3} d x + \int \frac{2}{7 (x + 2)} d x)$

Step 5

Let $u_{1} = 2 x - 3$ . 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 $u_{1} = 2 x - 3$ . Find $\frac{d u_{1}}{d x}$ .

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Step 5.1.1

Differentiate $2 x - 3$ .

$\frac{d}{d x} [2 x - 3]$

Step 5.1.2

By the Sum Rule, the derivative of $2 x - 3$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 3]$ .

$\frac{d}{d x} [2 x] + \frac{d}{d x} [- 3]$

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

$2 \frac{d}{d x} [x] + \frac{d}{d x} [- 3]$

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 $n = 1$ .

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

Step 5.1.3.3

Multiply $2$ by $1$ .

$2 + \frac{d}{d x} [- 3]$

Step 5.1.4

Differentiate using the Constant Rule.

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Step 5.1.4.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ 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 $d u_{1}$ .

$7 (\frac{3}{7} \int \frac{1}{u_{1}} \cdot \frac{1}{2} d u_{1} + \int \frac{2}{7 (x + 2)} d x)$

Step 6

Simplify.

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Step 6.1

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

$7 (\frac{3}{7} \int \frac{1}{u_{1} \cdot 2} d u_{1} + \int \frac{2}{7 (x + 2)} d x)$

Step 6.2

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

$7 (\frac{3}{7} \int \frac{1}{2 u_{1}} d u_{1} + \int \frac{2}{7 (x + 2)} d x)$

Step 7

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

$7 (\frac{3}{7} (\frac{1}{2} \int \frac{1}{u_{1}} d u_{1}) + \int \frac{2}{7 (x + 2)} d x)$

Step 8

Simplify.

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Step 8.1

Multiply $\frac{1}{2}$ by $\frac{3}{7}$ .

$7 (\frac{3}{2 \cdot 7} \int \frac{1}{u_{1}} d u_{1} + \int \frac{2}{7 (x + 2)} d x)$

Step 8.2

Multiply $2$ by $7$ .

$7 (\frac{3}{14} \int \frac{1}{u_{1}} d u_{1} + \int \frac{2}{7 (x + 2)} d x)$

Step 9

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

$7 (\frac{3}{14} (\ln (| u_{1} |) + C) + \int \frac{2}{7 (x + 2)} d x)$

Step 10

Since $\frac{2}{7}$ is constant with respect to $x$ , move $\frac{2}{7}$ out of the integral.

$7 (\frac{3}{14} (\ln (| u_{1} |) + C) + \frac{2}{7} \int \frac{1}{x + 2} d x)$

Step 11

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

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Step 11.1

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

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Step 11.1.1

Differentiate $x + 2$ .

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

Step 11.1.2

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

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

Step 11.1.3

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

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

Step 11.1.4

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

$1 + 0$

Step 11.1.5

Add $1$ and $0$ .

$1$

Step 11.2

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

$7 (\frac{3}{14} (\ln (| u_{1} |) + C) + \frac{2}{7} \int \frac{1}{u_{2}} d u_{2})$

Step 12

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

$7 (\frac{3}{14} (\ln (| u_{1} |) + C) + \frac{2}{7} (\ln (| u_{2} |) + C))$

Step 13

Simplify.

$7 (\frac{3}{14} \ln (| u_{1} |) + \frac{2}{7} \ln (| u_{2} |)) + C$

Step 14

Substitute back in for each integration substitution variable.

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Step 14.1

Replace all occurrences of $u_{1}$ with $2 x - 3$ .

$7 (\frac{3}{14} \ln (| 2 x - 3 |) + \frac{2}{7} \ln (| u_{2} |)) + C$

Step 14.2

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

$7 (\frac{3}{14} \ln (| 2 x - 3 |) + \frac{2}{7} \ln (| x + 2 |)) + C$

Step 15

Simplify.

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Step 15.1

Simplify each term.

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Step 15.1.1

Combine $\frac{3}{14}$ and $\ln (| 2 x - 3 |)$ .

$7 (\frac{3 \ln (| 2 x - 3 |)}{14} + \frac{2}{7} \ln (| x + 2 |)) + C$

Step 15.1.2

Combine $\frac{2}{7}$ and $\ln (| x + 2 |)$ .

$7 (\frac{3 \ln (| 2 x - 3 |)}{14} + \frac{2 \ln (| x + 2 |)}{7}) + C$

Step 15.2

To write $\frac{2 \ln (| x + 2 |)}{7}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$7 (\frac{3 \ln (| 2 x - 3 |)}{14} + \frac{2 \ln (| x + 2 |)}{7} \cdot \frac{2}{2}) + C$

Step 15.3

Write each expression with a common denominator of $14$ , by multiplying each by an appropriate factor of $1$ .

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Step 15.3.1

Multiply $\frac{2 \ln (| x + 2 |)}{7}$ by $\frac{2}{2}$ .

$7 (\frac{3 \ln (| 2 x - 3 |)}{14} + \frac{2 \ln (| x + 2 |) \cdot 2}{7 \cdot 2}) + C$

Step 15.3.2

Multiply $7$ by $2$ .

$7 (\frac{3 \ln (| 2 x - 3 |)}{14} + \frac{2 \ln (| x + 2 |) \cdot 2}{14}) + C$

Step 15.4

Combine the numerators over the common denominator.

$7 \frac{3 \ln (| 2 x - 3 |) + 2 \ln (| x + 2 |) \cdot 2}{14} + C$

Step 15.5

Cancel the common factor of $7$ .

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Step 15.5.1

Factor $7$ out of $14$ .

$7 \frac{3 \ln (| 2 x - 3 |) + 2 \ln (| x + 2 |) \cdot 2}{7 (2)} + C$

Step 15.5.2

Cancel the common factor.

$7 \frac{3 \ln (| 2 x - 3 |) + 2 \ln (| x + 2 |) \cdot 2}{7 \cdot 2} + C$

Step 15.5.3

Rewrite the expression.

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

Step 15.6

Multiply $2$ by $2$ .

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

Step 16

Reorder terms.

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