Calculus Examples

$\frac{2 x - 13}{(x + 1) (x - 2)}$

Step 1

Write $\frac{2 x - 13}{(x + 1) (x - 2)}$ as a function.

$f (x) = \frac{2 x - 13}{(x + 1) (x - 2)}$

Step 2

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

$F (x) = \int \frac{2 x - 13}{(x + 1) (x - 2)} d x$

Step 4

Write the fraction using partial fraction decomposition.

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Step 4.1

Decompose the fraction and multiply through by the common denominator.

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Step 4.1.1

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 $A$ .

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

Step 4.1.2

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

Step 4.1.3

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

$\frac{(2 x - 13) (x + 1) (x - 2)}{(x + 1) (x - 2)} = \frac{(A) (x + 1) (x - 2)}{x + 1} + \frac{(B) (x + 1) (x - 2)}{x - 2}$

Step 4.1.4

Cancel the common factor of $x + 1$ .

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Step 4.1.4.1

Cancel the common factor.

$\frac{(2 x - 13) (x + 1) (x - 2)}{(x + 1) (x - 2)} = \frac{(A) (x + 1) (x - 2)}{x + 1} + \frac{(B) (x + 1) (x - 2)}{x - 2}$

Step 4.1.4.2

Rewrite the expression.

$\frac{(2 x - 13) (x - 2)}{x - 2} = \frac{(A) (x + 1) (x - 2)}{x + 1} + \frac{(B) (x + 1) (x - 2)}{x - 2}$

Step 4.1.5

Cancel the common factor of $x - 2$ .

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Step 4.1.5.1

Cancel the common factor.

$\frac{(2 x - 13) (x - 2)}{x - 2} = \frac{(A) (x + 1) (x - 2)}{x + 1} + \frac{(B) (x + 1) (x - 2)}{x - 2}$

Step 4.1.5.2

Divide $2 x - 13$ by $1$ .

$2 x - 13 = \frac{(A) (x + 1) (x - 2)}{x + 1} + \frac{(B) (x + 1) (x - 2)}{x - 2}$

Step 4.1.6

Simplify each term.

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Step 4.1.6.1

Cancel the common factor of $x + 1$ .

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Step 4.1.6.1.1

Cancel the common factor.

$2 x - 13 = \frac{A (x + 1) (x - 2)}{x + 1} + \frac{(B) (x + 1) (x - 2)}{x - 2}$

Step 4.1.6.1.2

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

$2 x - 13 = (A) (x - 2) + \frac{(B) (x + 1) (x - 2)}{x - 2}$

Step 4.1.6.2

Apply the distributive property.

$2 x - 13 = A x + A \cdot - 2 + \frac{(B) (x + 1) (x - 2)}{x - 2}$

Step 4.1.6.3

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

$2 x - 13 = A x - 2 \cdot A + \frac{(B) (x + 1) (x - 2)}{x - 2}$

Step 4.1.6.4

Cancel the common factor of $x - 2$ .

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Step 4.1.6.4.1

Cancel the common factor.

$2 x - 13 = A x - 2 A + \frac{(B) (x + 1) (x - 2)}{x - 2}$

Step 4.1.6.4.2

Divide $(B) (x + 1)$ by $1$ .

$2 x - 13 = A x - 2 A + (B) (x + 1)$

Step 4.1.6.5

Apply the distributive property.

$2 x - 13 = A x - 2 A + B x + B \cdot 1$

Step 4.1.6.6

Multiply $B$ by $1$ .

$2 x - 13 = A x - 2 A + B x + B$

Step 4.1.7

Move $- 2 A$ .

$2 x - 13 = A x + B x - 2 A + B$

Step 4.2

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

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Step 4.2.1

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.

$2 = A + B$

Step 4.2.2

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.

$- 13 = - 2 A + B$

Step 4.2.3

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

$2 = A + B$

$- 13 = - 2 A + B$

$2 = A + B$

$- 13 = - 2 A + B$

Step 4.3

Solve the system of equations.

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Step 4.3.1

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

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Step 4.3.1.1

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

$A + B = 2$

$- 13 = - 2 A + B$

Step 4.3.1.2

Subtract $B$ from both sides of the equation.

$A = 2 - B$

$- 13 = - 2 A + B$

$A = 2 - B$

$- 13 = - 2 A + B$

Step 4.3.2

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

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Step 4.3.2.1

Replace all occurrences of $A$ in $- 13 = - 2 A + B$ with $2 - B$ .

$- 13 = - 2 (2 - B) + B$

$A = 2 - B$

Step 4.3.2.2

Simplify the right side.

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Step 4.3.2.2.1

Simplify $- 2 (2 - B) + B$ .

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Step 4.3.2.2.1.1

Simplify each term.

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Step 4.3.2.2.1.1.1

Apply the distributive property.

$- 13 = - 2 \cdot 2 - 2 (- B) + B$

$A = 2 - B$

Step 4.3.2.2.1.1.2

Multiply $- 2$ by $2$ .

$- 13 = - 4 - 2 (- B) + B$

$A = 2 - B$

Step 4.3.2.2.1.1.3

Multiply $- 1$ by $- 2$ .

$- 13 = - 4 + 2 B + B$

$A = 2 - B$

$- 13 = - 4 + 2 B + B$

$A = 2 - B$

Step 4.3.2.2.1.2

Add $2 B$ and $B$ .

$- 13 = - 4 + 3 B$

$A = 2 - B$

$- 13 = - 4 + 3 B$

$A = 2 - B$

$- 13 = - 4 + 3 B$

$A = 2 - B$

$- 13 = - 4 + 3 B$

$A = 2 - B$

Step 4.3.3

Solve for $B$ in $- 13 = - 4 + 3 B$ .

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Step 4.3.3.1

Rewrite the equation as $- 4 + 3 B = - 13$ .

$- 4 + 3 B = - 13$

$A = 2 - B$

Step 4.3.3.2

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

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Step 4.3.3.2.1

Add $4$ to both sides of the equation.

$3 B = - 13 + 4$

$A = 2 - B$

Step 4.3.3.2.2

Add $- 13$ and $4$ .

$3 B = - 9$

$A = 2 - B$

$3 B = - 9$

$A = 2 - B$

Step 4.3.3.3

Divide each term in $3 B = - 9$ by $3$ and simplify.

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Step 4.3.3.3.1

Divide each term in $3 B = - 9$ by $3$ .

$\frac{3 B}{3} = \frac{- 9}{3}$

$A = 2 - B$

Step 4.3.3.3.2

Simplify the left side.

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Step 4.3.3.3.2.1

Cancel the common factor of $3$ .

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Step 4.3.3.3.2.1.1

Cancel the common factor.

$\frac{3 B}{3} = \frac{- 9}{3}$

$A = 2 - B$

Step 4.3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{- 9}{3}$

$A = 2 - B$

$B = \frac{- 9}{3}$

$A = 2 - B$

$B = \frac{- 9}{3}$

$A = 2 - B$

Step 4.3.3.3.3

Simplify the right side.

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Step 4.3.3.3.3.1

Divide $- 9$ by $3$ .

$B = - 3$

$A = 2 - B$

$B = - 3$

$A = 2 - B$

$B = - 3$

$A = 2 - B$

$B = - 3$

$A = 2 - B$

Step 4.3.4

Replace all occurrences of $B$ with $- 3$ in each equation.

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Step 4.3.4.1

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

$A = 2 - (- 3)$

$B = - 3$

Step 4.3.4.2

Simplify the right side.

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Step 4.3.4.2.1

Simplify $2 - (- 3)$ .

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Step 4.3.4.2.1.1

Multiply $- 1$ by $- 3$ .

$A = 2 + 3$

$B = - 3$

Step 4.3.4.2.1.2

Add $2$ and $3$ .

$A = 5$

$B = - 3$

$A = 5$

$B = - 3$

$A = 5$

$B = - 3$

$A = 5$

$B = - 3$

Step 4.3.5

List all of the solutions.

$A = 5, B = - 3$

Step 4.4

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

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

Step 4.5

Move the negative in front of the fraction.

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

Step 5

Split the single integral into multiple integrals.

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

Step 6

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

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

Step 7

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

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Step 7.1

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

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Step 7.1.1

Differentiate $x + 1$ .

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

Step 7.1.2

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

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

Step 7.1.3

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

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

Step 7.1.4

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

$1 + 0$

Step 7.1.5

Add $1$ and $0$ .

$1$

Step 7.2

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

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

Step 8

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

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

Step 9

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

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

Step 10

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

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

Step 11

Multiply $3$ by $- 1$ .

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

Step 12

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

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Step 12.1

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

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Step 12.1.1

Differentiate $x - 2$ .

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

Step 12.1.2

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 12.1.3

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 12.1.4

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

$1 + 0$

Step 12.1.5

Add $1$ and $0$ .

$1$

Step 12.2

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

$5 (\ln (| u_{1} |) + C) - 3 \int \frac{1}{u_{2}} d u_{2}$

Step 13

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

$5 (\ln (| u_{1} |) + C) - 3 (\ln (| u_{2} |) + C)$

Step 14

Simplify.

$5 \ln (| u_{1} |) - 3 \ln (| u_{2} |) + C$

Step 15

Substitute back in for each integration substitution variable.

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Step 15.1

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

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

Step 15.2

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

$5 \ln (| x + 1 |) - 3 \ln (| x - 2 |) + C$

Step 16

The answer is the antiderivative of the function $f (x) = \frac{2 x - 13}{(x + 1) (x - 2)}$ .

$F (x) =$ $5 \ln (| x + 1 |) - 3 \ln (| x - 2 |) + C$