Calculus Examples

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

Step 1

Write $\frac{1}{x^{2} - 1}$ as a function.

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

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.

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

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

Factor the fraction.

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Step 4.1.1.1

Rewrite $1$ as $1^{2}$ .

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

Step 4.1.1.2

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 $b = 1$ .

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

Step 4.1.2

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 4.1.3

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

Step 4.1.4

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

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

Step 4.1.5

Cancel the common factor of $x + 1$ .

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Step 4.1.5.1

Cancel the common factor.

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

Step 4.1.5.2

Rewrite the expression.

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

Step 4.1.6

Cancel the common factor of $x - 1$ .

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Step 4.1.6.1

Cancel the common factor.

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

Step 4.1.6.2

Rewrite the expression.

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

Step 4.1.7

Simplify each term.

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Step 4.1.7.1

Cancel the common factor of $x + 1$ .

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Step 4.1.7.1.1

Cancel the common factor.

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

Step 4.1.7.1.2

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

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

Step 4.1.7.2

Apply the distributive property.

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

Step 4.1.7.3

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

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

Step 4.1.7.4

Rewrite $- 1 A$ as $- A$ .

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

Step 4.1.7.5

Cancel the common factor of $x - 1$ .

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Step 4.1.7.5.1

Cancel the common factor.

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

Step 4.1.7.5.2

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

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

Step 4.1.7.6

Apply the distributive property.

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

Step 4.1.7.7

Multiply $B$ by $1$ .

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

Step 4.1.8

Move $- A$ .

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

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.

$0 = A + B$

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.

$1 = - 1 A + B$

Step 4.2.3

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

$0 = A + B$

$1 = - 1 A + B$

$0 = A + B$

$1 = - 1 A + B$

Step 4.3

Solve the system of equations.

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Step 4.3.1

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

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Step 4.3.1.1

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

$A + B = 0$

$1 = - 1 A + B$

Step 4.3.1.2

Subtract $B$ from both sides of the equation.

$A = - B$

$1 = - 1 A + B$

$A = - B$

$1 = - 1 A + B$

Step 4.3.2

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

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Step 4.3.2.1

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

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

$A = - B$

Step 4.3.2.2

Simplify the right side.

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Step 4.3.2.2.1

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

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Step 4.3.2.2.1.1

Multiply $- 1 (- B)$ .

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Step 4.3.2.2.1.1.1

Multiply $- 1$ by $- 1$ .

$1 = 1 B + B$

$A = - B$

Step 4.3.2.2.1.1.2

Multiply $B$ by $1$ .

$1 = B + B$

$A = - B$

$1 = B + B$

$A = - B$

Step 4.3.2.2.1.2

Add $B$ and $B$ .

$1 = 2 B$

$A = - B$

$1 = 2 B$

$A = - B$

$1 = 2 B$

$A = - B$

$1 = 2 B$

$A = - B$

Step 4.3.3

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

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Step 4.3.3.1

Rewrite the equation as $2 B = 1$ .

$2 B = 1$

$A = - B$

Step 4.3.3.2

Divide each term in $2 B = 1$ by $2$ and simplify.

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Step 4.3.3.2.1

Divide each term in $2 B = 1$ by $2$ .

$\frac{2 B}{2} = \frac{1}{2}$

$A = - B$

Step 4.3.3.2.2

Simplify the left side.

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Step 4.3.3.2.2.1

Cancel the common factor of $2$ .

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Step 4.3.3.2.2.1.1

Cancel the common factor.

$\frac{2 B}{2} = \frac{1}{2}$

$A = - B$

Step 4.3.3.2.2.1.2

Divide $B$ by $1$ .

$B = \frac{1}{2}$

$A = - B$

$B = \frac{1}{2}$

$A = - B$

$B = \frac{1}{2}$

$A = - B$

$B = \frac{1}{2}$

$A = - B$

$B = \frac{1}{2}$

$A = - B$

Step 4.3.4

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

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Step 4.3.4.1

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

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

$B = \frac{1}{2}$

Step 4.3.4.2

Simplify the right side.

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Step 4.3.4.2.1

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

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

$B = \frac{1}{2}$

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

$B = \frac{1}{2}$

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

$B = \frac{1}{2}$

Step 4.3.5

List all of the solutions.

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

Step 4.4

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

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

Step 4.5

Simplify.

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Step 4.5.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 4.5.2

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

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

Step 4.5.3

Move $2$ to the left of $x + 1$ .

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

Step 4.5.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 4.5.5

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

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

Step 5

Split the single integral into multiple integrals.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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Step 8.1

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

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Step 8.1.1

Differentiate $x + 1$ .

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

Step 8.1.2

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 8.1.3

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 8.1.4

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

$1 + 0$

Step 8.1.5

Add $1$ and $0$ .

$1$

Step 8.2

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

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

Step 9

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

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

Step 10

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

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

Step 11

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

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Step 11.1

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

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Step 11.1.1

Differentiate $x - 1$ .

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

Step 11.1.2

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 11.1.3

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

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

Step 11.1.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ 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 $d u_{2}$ .

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

Step 12

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

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

Step 13

Simplify.

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

Step 14

Substitute back in for each integration substitution variable.

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Step 14.1

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

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

Step 14.2

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

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

Step 15

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

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