Calculus Examples

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

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

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

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

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

Step 1.1.5.4

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

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

Step 1.1.5.5

Cancel the common factor of $x - 1$ .

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Step 1.1.5.5.1

Cancel the common factor.

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

Step 1.1.5.5.2

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.

$1 = - 1 A$

Step 1.2.3

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

$0 = A + B$

$1 = - 1 A$

$0 = A + B$

$1 = - 1 A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $A$ in $1 = - 1 A$ .

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Step 1.3.1.1

Rewrite the equation as $- 1 A = 1$ .

$- 1 A = 1$

$0 = A + B$

Step 1.3.1.2

Divide each term in $- 1 A = 1$ by $- 1$ and simplify.

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Step 1.3.1.2.1

Divide each term in $- 1 A = 1$ by $- 1$ .

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

$0 = A + B$

Step 1.3.1.2.2

Simplify the left side.

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Step 1.3.1.2.2.1

Dividing two negative values results in a positive value.

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

$0 = A + B$

Step 1.3.1.2.2.2

Divide $A$ by $1$ .

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

$0 = A + B$

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

$0 = A + B$

Step 1.3.1.2.3

Simplify the right side.

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Step 1.3.1.2.3.1

Divide $1$ by $- 1$ .

$A = - 1$

$0 = A + B$

$A = - 1$

$0 = A + B$

$A = - 1$

$0 = A + B$

$A = - 1$

$0 = A + B$

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

Add $1$ to 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

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

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

Step 4

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

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

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

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

$- \ln (| x |) + \ln (| x - 1 |) + C$