Calculus Examples

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

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

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 $B$ .

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

Step 1.1.2

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

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

Step 1.1.3

Cancel the common factor of $x$ .

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Step 1.1.3.1

Cancel the common factor.

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

Step 1.1.3.2

Rewrite the expression.

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

Step 1.1.4

Cancel the common factor of $x + 1$ .

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Step 1.1.4.1

Cancel the common factor.

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

Step 1.1.4.2

Rewrite the expression.

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

Step 1.1.5

Simplify each term.

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Step 1.1.5.1

Cancel the common factor of $x$ .

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Step 1.1.5.1.1

Cancel the common factor.

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

Step 1.1.5.1.2

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

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

Step 1.1.5.2

Apply the distributive property.

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

Step 1.1.5.3

Multiply $A$ by $1$ .

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

Step 1.1.5.4

Cancel the common factor of $x + 1$ .

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Step 1.1.5.4.1

Cancel the common factor.

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

Step 1.1.5.4.2

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

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

$1 = A x + A + B x$

Step 1.1.6

Move $A$ .

$1 = A x + B x + A$

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.

$0 = A + B$

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.

$1 = A$

Step 1.2.3

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

$0 = A + B$

$1 = A$

$0 = A + B$

$1 = A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Rewrite the equation as $A = 1$ .

$A = 1$

$0 = A + B$

Step 1.3.2

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

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Step 1.3.2.1

Replace all occurrences of $A$ in $0 = A + B$ with $1$ .

$0 = (1) + B$

$A = 1$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Remove parentheses.

$0 = 1 + B$

$A = 1$

$0 = 1 + B$

$A = 1$

$0 = 1 + B$

$A = 1$

Step 1.3.3

Solve for $B$ in $0 = 1 + B$ .

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Step 1.3.3.1

Rewrite the equation as $1 + B = 0$ .

$1 + B = 0$

$A = 1$

Step 1.3.3.2

Subtract $1$ from both sides of the equation.

$B = - 1$

$A = 1$

$B = - 1$

$A = 1$

Step 1.3.4

Solve the system of equations.

$B = - 1 A = 1$

Step 1.3.5

List all of the solutions.

$B = - 1, A = 1$

Step 1.4

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

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

Step 1.5

Move the negative in front of the fraction.

$\int \frac{1}{x} - \frac{1}{x + 1} d x$

Step 2

Split the single integral into multiple integrals.

$\int \frac{1}{x} d x + \int - \frac{1}{x + 1} d x$

Step 3

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

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

Step 4

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

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

Step 5

Let $u = x + 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1

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

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Step 5.1.1

Differentiate $x + 1$ .

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

Step 5.1.2

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 5.1.3

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 5.1.4

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

$1 + 0$

Step 5.1.5

Add $1$ and $0$ .

$1$

Step 5.2

Rewrite the problem using $u$ and $d u$ .

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

Step 6

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

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

Step 7

Simplify.

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

Step 8

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| x |}{| u |}) + C$

Step 9

Replace all occurrences of $u$ with $x + 1$ .

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