Calculus Examples

$\frac{4}{x^{2} - 4}$

Step 1

Write $\frac{4}{x^{2} - 4}$ as a function.

$f (x) = \frac{4}{x^{2} - 4}$

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.

$F (x) = \int \frac{4}{x^{2} - 4} d x$

Step 4

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

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

Step 5

Write the fraction using partial fraction decomposition.

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Step 5.1

Decompose the fraction and multiply through by the common denominator.

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Step 5.1.1

Factor the fraction.

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Step 5.1.1.1

Rewrite $4$ as $2^{2}$ .

$\frac{1}{x^{2} - 2^{2}}$

Step 5.1.1.2

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

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

Step 5.1.2

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 $A$ .

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

Step 5.1.3

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

Step 5.1.4

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

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

Step 5.1.5

Cancel the common factor of $x + 2$ .

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Step 5.1.5.1

Cancel the common factor.

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

Step 5.1.5.2

Rewrite the expression.

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

Step 5.1.6

Cancel the common factor of $x - 2$ .

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Step 5.1.6.1

Cancel the common factor.

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

Step 5.1.6.2

Rewrite the expression.

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

Step 5.1.7

Simplify each term.

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Step 5.1.7.1

Cancel the common factor of $x + 2$ .

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Step 5.1.7.1.1

Cancel the common factor.

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

Step 5.1.7.1.2

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

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

Step 5.1.7.2

Apply the distributive property.

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

Step 5.1.7.3

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

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

Step 5.1.7.4

Cancel the common factor of $x - 2$ .

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Step 5.1.7.4.1

Cancel the common factor.

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

Step 5.1.7.4.2

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

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

Step 5.1.7.5

Apply the distributive property.

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

Step 5.1.7.6

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

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

Step 5.1.8

Move $- 2 A$ .

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

Step 5.2

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

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Step 5.2.1

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.

$0 = A + B$

Step 5.2.2

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.

$1 = - 2 A + 2 B$

Step 5.2.3

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

$0 = A + B$

$1 = - 2 A + 2 B$

$0 = A + B$

$1 = - 2 A + 2 B$

Step 5.3

Solve the system of equations.

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Step 5.3.1

Solve for $A$ in $0 = A + B$ .

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Step 5.3.1.1

Rewrite the equation as $A + B = 0$ .

$A + B = 0$

$1 = - 2 A + 2 B$

Step 5.3.1.2

Subtract $B$ from both sides of the equation.

$A = - B$

$1 = - 2 A + 2 B$

$A = - B$

$1 = - 2 A + 2 B$

Step 5.3.2

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

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Step 5.3.2.1

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

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

$A = - B$

Step 5.3.2.2

Simplify the right side.

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Step 5.3.2.2.1

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

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Step 5.3.2.2.1.1

Multiply $- 1$ by $- 2$ .

$1 = 2 B + 2 B$

$A = - B$

Step 5.3.2.2.1.2

Add $2 B$ and $2 B$ .

$1 = 4 B$

$A = - B$

$1 = 4 B$

$A = - B$

$1 = 4 B$

$A = - B$

$1 = 4 B$

$A = - B$

Step 5.3.3

Solve for $B$ in $1 = 4 B$ .

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Step 5.3.3.1

Rewrite the equation as $4 B = 1$ .

$4 B = 1$

$A = - B$

Step 5.3.3.2

Divide each term in $4 B = 1$ by $4$ and simplify.

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Step 5.3.3.2.1

Divide each term in $4 B = 1$ by $4$ .

$\frac{4 B}{4} = \frac{1}{4}$

$A = - B$

Step 5.3.3.2.2

Simplify the left side.

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Step 5.3.3.2.2.1

Cancel the common factor of $4$ .

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Step 5.3.3.2.2.1.1

Cancel the common factor.

$\frac{4 B}{4} = \frac{1}{4}$

$A = - B$

Step 5.3.3.2.2.1.2

Divide $B$ by $1$ .

$B = \frac{1}{4}$

$A = - B$

$B = \frac{1}{4}$

$A = - B$

$B = \frac{1}{4}$

$A = - B$

$B = \frac{1}{4}$

$A = - B$

$B = \frac{1}{4}$

$A = - B$

Step 5.3.4

Replace all occurrences of $B$ with $\frac{1}{4}$ in each equation.

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Step 5.3.4.1

Replace all occurrences of $B$ in $A = - B$ with $\frac{1}{4}$ .

$A = - (\frac{1}{4})$

$B = \frac{1}{4}$

Step 5.3.4.2

Simplify the right side.

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Step 5.3.4.2.1

Multiply $- 1$ by $\frac{1}{4}$ .

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

$B = \frac{1}{4}$

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

$B = \frac{1}{4}$

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

$B = \frac{1}{4}$

Step 5.3.5

List all of the solutions.

$A = - \frac{1}{4}, B = \frac{1}{4}$

Step 5.4

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

$\frac{- \frac{1}{4}}{x + 2} + \frac{\frac{1}{4}}{x - 2}$

Step 5.5

Simplify.

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Step 5.5.1

Multiply the numerator by the reciprocal of the denominator.

$- \frac{1}{4} \cdot \frac{1}{x + 2} + \frac{\frac{1}{4}}{x - 2}$

Step 5.5.2

Multiply $\frac{1}{x + 2}$ by $\frac{1}{4}$ .

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

Step 5.5.3

Move $4$ to the left of $x + 2$ .

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

Step 5.5.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.5.5

Multiply $\frac{1}{4}$ by $\frac{1}{x - 2}$ .

$4 \int - \frac{1}{4 (x + 2)} + \frac{1}{4 (x - 2)} d x$

Step 6

Split the single integral into multiple integrals.

$4 (\int - \frac{1}{4 (x + 2)} d x + \int \frac{1}{4 (x - 2)} d x)$

Step 7

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

$4 (- \int \frac{1}{4 (x + 2)} d x + \int \frac{1}{4 (x - 2)} d x)$

Step 8

Since $\frac{1}{4}$ is constant with respect to $x$ , move $\frac{1}{4}$ out of the integral.

$4 (- (\frac{1}{4} \int \frac{1}{x + 2} d x) + \int \frac{1}{4 (x - 2)} d x)$

Step 9

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

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Step 9.1

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

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Step 9.1.1

Differentiate $x + 2$ .

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

Step 9.1.2

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 9.1.3

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 9.1.4

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

$1 + 0$

Step 9.1.5

Add $1$ and $0$ .

$1$

Step 9.2

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

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

Step 10

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

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

Step 11

Since $\frac{1}{4}$ is constant with respect to $x$ , move $\frac{1}{4}$ out of the integral.

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

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 $d u_{2}$ .

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

Step 13

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

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

Step 14

Simplify.

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

Step 15

Substitute back in for each integration substitution variable.

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Step 15.1

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

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

Step 15.2

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

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

Step 16

Simplify.

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Step 16.1

Simplify each term.

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Step 16.1.1

Combine $\ln (| x + 2 |)$ and $\frac{1}{4}$ .

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

Step 16.1.2

Combine $\frac{1}{4}$ and $\ln (| x - 2 |)$ .

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

Step 16.2

Combine the numerators over the common denominator.

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

Step 16.3

Cancel the common factor of $4$ .

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Step 16.3.1

Cancel the common factor.

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

Step 16.3.2

Rewrite the expression.

$- \ln (| x + 2 |) + \ln (| x - 2 |) + C$

Step 17

The answer is the antiderivative of the function $f (x) = \frac{4}{x^{2} - 4}$ .

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