Calculus Examples

$\int \frac{x + 4}{x^{2} + 5 x - 6} d x$

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 $x^{2} + 5 x - 6$ using the AC method.

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Step 1.1.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $5$ .

$- 1, 6$

Step 1.1.1.2

Write the factored form using these integers.

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

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 $A$ .

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

Step 1.1.3

$\frac{A}{x - 1} + \frac{B}{x + 6}$

Step 1.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(x - 1) (x + 6)$ .

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

Step 1.1.5

Cancel the common factor of $x - 1$ .

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Step 1.1.5.1

Cancel the common factor.

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

Step 1.1.5.2

Rewrite the expression.

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

Step 1.1.6

Cancel the common factor of $x + 6$ .

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Step 1.1.6.1

Cancel the common factor.

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

Step 1.1.6.2

Divide $x + 4$ by $1$ .

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

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

Cancel the common factor of $x - 1$ .

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Step 1.1.7.1.1

Cancel the common factor.

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

Step 1.1.7.1.2

Divide $(A) (x + 6)$ by $1$ .

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

Step 1.1.7.2

Apply the distributive property.

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

Step 1.1.7.3

Move $6$ to the left of $A$ .

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

Step 1.1.7.4

Cancel the common factor of $x + 6$ .

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Step 1.1.7.4.1

Cancel the common factor.

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

Step 1.1.7.4.2

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

$x + 4 = A x + 6 A + (B) (x - 1)$

Step 1.1.7.5

Apply the distributive property.

$x + 4 = A x + 6 A + B x + B \cdot - 1$

Step 1.1.7.6

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

$x + 4 = A x + 6 A + B x - 1 \cdot B$

Step 1.1.7.7

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

$x + 4 = A x + 6 A + B x - B$

Step 1.1.8

Move $6 A$ .

$x + 4 = A x + B x + 6 A - B$

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.

$1 = A + B$

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.

$4 = 6 A - 1 B$

Step 1.2.3

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

$1 = A + B$

$4 = 6 A - 1 B$

$1 = A + B$

$4 = 6 A - 1 B$

Step 1.3

Solve the system of equations.

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Step 1.3.1

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

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Step 1.3.1.1

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

$A + B = 1$

$4 = 6 A - 1 B$

Step 1.3.1.2

Subtract $B$ from both sides of the equation.

$A = 1 - B$

$4 = 6 A - 1 B$

$A = 1 - B$

$4 = 6 A - 1 B$

Step 1.3.2

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

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Step 1.3.2.1

Replace all occurrences of $A$ in $4 = 6 A - 1 B$ with $1 - B$ .

$4 = 6 (1 - B) - 1 B$

$A = 1 - B$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Simplify $6 (1 - B) - 1 B$ .

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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.

$4 = 6 \cdot 1 + 6 (- B) - 1 B$

$A = 1 - B$

Step 1.3.2.2.1.1.2

Multiply $6$ by $1$ .

$4 = 6 + 6 (- B) - 1 B$

$A = 1 - B$

Step 1.3.2.2.1.1.3

Multiply $- 1$ by $6$ .

$4 = 6 - 6 B - 1 B$

$A = 1 - B$

Step 1.3.2.2.1.1.4

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

$4 = 6 - 6 B - B$

$A = 1 - B$

$4 = 6 - 6 B - B$

$A = 1 - B$

Step 1.3.2.2.1.2

Subtract $B$ from $- 6 B$ .

$4 = 6 - 7 B$

$A = 1 - B$

$4 = 6 - 7 B$

$A = 1 - B$

$4 = 6 - 7 B$

$A = 1 - B$

$4 = 6 - 7 B$

$A = 1 - B$

Step 1.3.3

Solve for $B$ in $4 = 6 - 7 B$ .

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Step 1.3.3.1

Rewrite the equation as $6 - 7 B = 4$ .

$6 - 7 B = 4$

$A = 1 - B$

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 $6$ from both sides of the equation.

$- 7 B = 4 - 6$

$A = 1 - B$

Step 1.3.3.2.2

Subtract $6$ from $4$ .

$- 7 B = - 2$

$A = 1 - B$

$- 7 B = - 2$

$A = 1 - B$

Step 1.3.3.3

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

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Step 1.3.3.3.1

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

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

$A = 1 - B$

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 $- 7$ .

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Step 1.3.3.3.2.1.1

Cancel the common factor.

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

$A = 1 - B$

Step 1.3.3.3.2.1.2

Divide $B$ by $1$ .

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

$A = 1 - B$

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

$A = 1 - B$

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

$A = 1 - B$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Dividing two negative values results in a positive value.

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

$A = 1 - B$

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

$A = 1 - B$

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

$A = 1 - B$

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

$A = 1 - B$

Step 1.3.4

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

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Step 1.3.4.1

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

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

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

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

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

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Step 1.3.4.2.1.1

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

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

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

Step 1.3.4.2.1.2

Combine the numerators over the common denominator.

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

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

Step 1.3.4.2.1.3

Subtract $2$ from $7$ .

$A = \frac{5}{7}$

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

$A = \frac{5}{7}$

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

$A = \frac{5}{7}$

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

$A = \frac{5}{7}$

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

Step 1.3.5

List all of the solutions.

$A = \frac{5}{7}, B = \frac{2}{7}$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{x - 1} + \frac{B}{x + 6}$ with the values found for $A$ and $B$ .

$\frac{\frac{5}{7}}{x - 1} + \frac{\frac{2}{7}}{x + 6}$

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{5}{7} \cdot \frac{1}{x - 1} + \frac{\frac{2}{7}}{x + 6}$

Step 1.5.2

Multiply $\frac{5}{7}$ by $\frac{1}{x - 1}$ .

$\frac{5}{7 (x - 1)} + \frac{\frac{2}{7}}{x + 6}$

Step 1.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{5}{7 (x - 1)} + \frac{2}{7} \cdot \frac{1}{x + 6}$

Step 1.5.4

Multiply $\frac{2}{7}$ by $\frac{1}{x + 6}$ .

$\int \frac{5}{7 (x - 1)} + \frac{2}{7 (x + 6)} d x$

Step 2

Split the single integral into multiple integrals.

$\int \frac{5}{7 (x - 1)} d x + \int \frac{2}{7 (x + 6)} d x$

Step 3

Since $\frac{5}{7}$ is constant with respect to $x$ , move $\frac{5}{7}$ out of the integral.

$\frac{5}{7} \int \frac{1}{x - 1} d x + \int \frac{2}{7 (x + 6)} d x$

Step 4

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

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Step 4.1

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

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Step 4.1.1

Differentiate $x - 1$ .

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

Step 4.1.2

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 4.1.3

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 4.1.4

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

$1 + 0$

Step 4.1.5

Add $1$ and $0$ .

$1$

Step 4.2

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

$\frac{5}{7} \int \frac{1}{u_{1}} d u_{1} + \int \frac{2}{7 (x + 6)} d x$

Step 5

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

$\frac{5}{7} (\ln (| u_{1} |) + C) + \int \frac{2}{7 (x + 6)} d x$

Step 6

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

$\frac{5}{7} (\ln (| u_{1} |) + C) + \frac{2}{7} \int \frac{1}{x + 6} d x$

Step 7

Let $u_{2} = x + 6$ . Then $d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 7.1

Let $u_{2} = x + 6$ . Find $\frac{d u_{2}}{d x}$ .

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Step 7.1.1

Differentiate $x + 6$ .

$\frac{d}{d x} [x + 6]$

Step 7.1.2

By the Sum Rule, the derivative of $x + 6$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [6]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [6]$

Step 7.1.3

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

$1 + \frac{d}{d x} [6]$

Step 7.1.4

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

$1 + 0$

Step 7.1.5

Add $1$ and $0$ .

$1$

Step 7.2

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

$\frac{5}{7} (\ln (| u_{1} |) + C) + \frac{2}{7} \int \frac{1}{u_{2}} d u_{2}$

Step 8

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

$\frac{5}{7} (\ln (| u_{1} |) + C) + \frac{2}{7} (\ln (| u_{2} |) + C)$

Step 9

Simplify.

$\frac{5}{7} \ln (| u_{1} |) + \frac{2}{7} \ln (| u_{2} |) + C$

Step 10

Substitute back in for each integration substitution variable.

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Step 10.1

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

$\frac{5}{7} \ln (| x - 1 |) + \frac{2}{7} \ln (| u_{2} |) + C$

Step 10.2

Replace all occurrences of $u_{2}$ with $x + 6$ .

$\frac{5}{7} \ln (| x - 1 |) + \frac{2}{7} \ln (| x + 6 |) + C$