Calculus Examples

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

Step 1

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

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

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.

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

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 $x$ out of $x^{2} - 2 x$ .

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Step 4.1.1.1

Factor $x$ out of $x^{2}$ .

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

Step 4.1.1.2

Factor $x$ out of $- 2 x$ .

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

Step 4.1.1.3

Factor $x$ out of $x \cdot x + x \cdot - 2$ .

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

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 $B$ .

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

Step 4.1.3

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

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

Step 4.1.4

Cancel the common factor of $x$ .

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Step 4.1.4.1

Cancel the common factor.

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

Step 4.1.4.2

Rewrite the expression.

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

Step 4.1.5

Cancel the common factor of $x - 2$ .

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Step 4.1.5.1

Cancel the common factor.

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

Step 4.1.5.2

Rewrite the expression.

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

Step 4.1.6

Simplify each term.

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Step 4.1.6.1

Cancel the common factor of $x$ .

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Step 4.1.6.1.1

Cancel the common factor.

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

Step 4.1.6.1.2

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

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

Step 4.1.6.2

Apply the distributive property.

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

Step 4.1.6.3

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

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

Step 4.1.6.4

Cancel the common factor of $x - 2$ .

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Step 4.1.6.4.1

Cancel the common factor.

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

Step 4.1.6.4.2

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

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

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

Step 4.1.7

Move $- 2 A$ .

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

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.

$1 = - 2 A$

Step 4.2.3

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

$0 = A + B$

$1 = - 2 A$

$0 = A + B$

$1 = - 2 A$

Step 4.3

Solve the system of equations.

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Step 4.3.1

Solve for $A$ in $1 = - 2 A$ .

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Step 4.3.1.1

Rewrite the equation as $- 2 A = 1$ .

$- 2 A = 1$

$0 = A + B$

Step 4.3.1.2

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

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Step 4.3.1.2.1

Divide each term in $- 2 A = 1$ by $- 2$ .

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

$0 = A + B$

Step 4.3.1.2.2

Simplify the left side.

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Step 4.3.1.2.2.1

Cancel the common factor of $- 2$ .

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Step 4.3.1.2.2.1.1

Cancel the common factor.

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

$0 = A + B$

Step 4.3.1.2.2.1.2

Divide $A$ by $1$ .

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

$0 = A + B$

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

$0 = A + B$

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

$0 = A + B$

Step 4.3.1.2.3

Simplify the right side.

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Step 4.3.1.2.3.1

Move the negative in front of the fraction.

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

$0 = A + B$

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

$0 = A + B$

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

$0 = A + B$

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

$0 = A + B$

Step 4.3.2

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

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Step 4.3.2.1

Replace all occurrences of $A$ in $0 = A + B$ with $- \frac{1}{2}$ .

$0 = (- \frac{1}{2}) + B$

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

Step 4.3.2.2

Simplify the right side.

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Step 4.3.2.2.1

Remove parentheses.

$0 = - \frac{1}{2} + B$

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

$0 = - \frac{1}{2} + B$

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

$0 = - \frac{1}{2} + B$

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

Step 4.3.3

Solve for $B$ in $0 = - \frac{1}{2} + B$ .

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Step 4.3.3.1

Rewrite the equation as $- \frac{1}{2} + B = 0$ .

$- \frac{1}{2} + B = 0$

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

Step 4.3.3.2

Add $\frac{1}{2}$ to both sides of the equation.

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

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

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

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

Step 4.3.4

Solve the system of equations.

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

Step 4.3.5

List all of the solutions.

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

Step 4.4

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

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

Step 4.5

Simplify.

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Step 4.5.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 4.5.2

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

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

Step 4.5.3

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

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

Step 4.5.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 4.5.5

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

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

Step 5

Split the single integral into multiple integrals.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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

Step 9

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

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

Step 10

Let $u = x - 2$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 10.1

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

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Step 10.1.1

Differentiate $x - 2$ .

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

Step 10.1.2

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 10.1.3

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 10.1.4

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

$1 + 0$

Step 10.1.5

Add $1$ and $0$ .

$1$

Step 10.2

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

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

Step 11

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

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

Step 12

Simplify.

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

Step 13

Replace all occurrences of $u$ with $x - 2$ .

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

Step 14

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

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