Calculus Examples

$\int_{2}^{3} \frac{1}{x^{2} - 4 x} d x$

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

Factor $x$ out of $x^{2} - 4 x$ .

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Step 1.1.1.1

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

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

Step 1.1.1.2

Factor $x$ out of $- 4 x$ .

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

Step 1.1.1.3

Factor $x$ out of $x \cdot x + x \cdot - 4$ .

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

Step 1.1.2

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 $B$ .

$\frac{A}{x} + \frac{B}{x - 4}$

Step 1.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $x (x - 4)$ .

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

Step 1.1.4

Cancel the common factor of $x$ .

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Step 1.1.4.1

Cancel the common factor.

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

Step 1.1.4.2

Rewrite the expression.

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

Step 1.1.5

Cancel the common factor of $x - 4$ .

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Step 1.1.5.1

Cancel the common factor.

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

Step 1.1.5.2

Rewrite the expression.

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

Step 1.1.6

Simplify each term.

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Step 1.1.6.1

Cancel the common factor of $x$ .

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Step 1.1.6.1.1

Cancel the common factor.

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

Step 1.1.6.1.2

Divide $A (x - 4)$ by $1$ .

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

Step 1.1.6.2

Apply the distributive property.

$1 = A x + A \cdot - 4 + \frac{(B) (x (x - 4))}{x - 4}$

Step 1.1.6.3

Move $- 4$ to the left of $A$ .

$1 = A x - 4 \cdot A + \frac{(B) (x (x - 4))}{x - 4}$

Step 1.1.6.4

Cancel the common factor of $x - 4$ .

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Step 1.1.6.4.1

Cancel the common factor.

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

Step 1.1.6.4.2

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

$1 = A x - 4 A + (B) (x)$

$1 = A x - 4 A + B x$

Step 1.1.7

Move $- 4 A$ .

$1 = A x + B x - 4 A$

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.

$1 = - 4 A$

Step 1.2.3

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

$0 = A + B$

$1 = - 4 A$

$0 = A + B$

$1 = - 4 A$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $A$ in $1 = - 4 A$ .

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Step 1.3.1.1

Rewrite the equation as $- 4 A = 1$ .

$- 4 A = 1$

$0 = A + B$

Step 1.3.1.2

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

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Step 1.3.1.2.1

Divide each term in $- 4 A = 1$ by $- 4$ .

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

$0 = A + B$

Step 1.3.1.2.2

Simplify the left side.

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Step 1.3.1.2.2.1

Cancel the common factor of $- 4$ .

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Step 1.3.1.2.2.1.1

Cancel the common factor.

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

$0 = A + B$

Step 1.3.1.2.2.1.2

Divide $A$ by $1$ .

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

$0 = A + B$

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

$0 = A + B$

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

$0 = A + B$

Step 1.3.1.2.3

Simplify the right side.

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Step 1.3.1.2.3.1

Move the negative in front of the fraction.

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

$0 = A + B$

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

$0 = A + B$

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

$0 = A + B$

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

$0 = A + B$

Step 1.3.2

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

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Step 1.3.2.1

Replace all occurrences of $A$ in $0 = A + B$ with $- \frac{1}{4}$ .

$0 = (- \frac{1}{4}) + B$

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

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Remove parentheses.

$0 = - \frac{1}{4} + B$

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

$0 = - \frac{1}{4} + B$

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

$0 = - \frac{1}{4} + B$

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

Step 1.3.3

Solve for $B$ in $0 = - \frac{1}{4} + B$ .

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Step 1.3.3.1

Rewrite the equation as $- \frac{1}{4} + B = 0$ .

$- \frac{1}{4} + B = 0$

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

Step 1.3.3.2

Add $\frac{1}{4}$ to both sides of the equation.

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

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

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

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

Step 1.3.4

Solve the system of equations.

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

Step 1.3.5

List all of the solutions.

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

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{x - 4}$ with the values found for $A$ and $B$ .

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

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.5.2

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

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

Step 1.5.3

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

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

Step 1.5.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.5.5

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

$\int_{2}^{3} - \frac{1}{4 x} + \frac{1}{4 (x - 4)} d x$

Step 2

Split the single integral into multiple integrals.

$\int_{2}^{3} - \frac{1}{4 x} d x + \int_{2}^{3} \frac{1}{4 (x - 4)} d x$

Step 3

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

$- \int_{2}^{3} \frac{1}{4 x} d x + \int_{2}^{3} \frac{1}{4 (x - 4)} d x$

Step 4

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

$- (\frac{1}{4} \int_{2}^{3} \frac{1}{x} d x) + \int_{2}^{3} \frac{1}{4 (x - 4)} d x$

Step 5

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

$- \frac{1}{4} \ln (| x |)]_{2}^{3} + \int_{2}^{3} \frac{1}{4 (x - 4)} d x$

Step 6

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

$- \frac{1}{4} \ln (| x |)]_{2}^{3} + \frac{1}{4} \int_{2}^{3} \frac{1}{x - 4} d x$

Step 7

Let $u = x - 4$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 7.1

Let $u = x - 4$ . Find $\frac{d u}{d x}$ .

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Step 7.1.1

Differentiate $x - 4$ .

$\frac{d}{d x} [x - 4]$

Step 7.1.2

By the Sum Rule, the derivative of $x - 4$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 4]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 4]$

Step 7.1.3

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

$1 + \frac{d}{d x} [- 4]$

Step 7.1.4

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$1 + 0$

Step 7.1.5

Add $1$ and $0$ .

$1$

Step 7.2

Substitute the lower limit in for $x$ in $u = x - 4$ .

$u_{lower} = 2 - 4$

Step 7.3

Subtract $4$ from $2$ .

$u_{lower} = - 2$

Step 7.4

Substitute the upper limit in for $x$ in $u = x - 4$ .

$u_{upper} = 3 - 4$

Step 7.5

Subtract $4$ from $3$ .

$u_{upper} = - 1$

Step 7.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = - 2$

$u_{upper} = - 1$

Step 7.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$- \frac{1}{4} \ln (| x |)]_{2}^{3} + \frac{1}{4} \int_{- 2}^{- 1} \frac{1}{u} d u$

Step 8

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

$- \frac{1}{4} \ln (| x |)]_{2}^{3} + \frac{1}{4} \ln (| u |)]_{- 2}^{- 1}$

Step 9

Substitute and simplify.

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Step 9.1

Evaluate $\ln (| x |)$ at $3$ and at $2$ .

$- \frac{1}{4} ((\ln (| 3 |)) - \ln (| 2 |)) + \frac{1}{4} \ln (| u |)]_{- 2}^{- 1}$

Step 9.2

Evaluate $\ln (| u |)$ at $- 1$ and at $- 2$ .

$- \frac{1}{4} ((\ln (| 3 |)) - \ln (| 2 |)) + \frac{1}{4} ((\ln (| - 1 |)) - \ln (| - 2 |))$

Step 9.3

Remove parentheses.

$- \frac{1}{4} (\ln (| 3 |) - \ln (| 2 |)) + \frac{1}{4} (\ln (| - 1 |) - \ln (| - 2 |))$

Step 10

Simplify.

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Step 10.1

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

$- \frac{1}{4} \ln (\frac{| 3 |}{| 2 |}) + \frac{1}{4} (\ln (| - 1 |) - \ln (| - 2 |))$

Step 10.2

Combine $\ln (\frac{| 3 |}{| 2 |})$ and $\frac{1}{4}$ .

$- \frac{\ln (\frac{| 3 |}{| 2 |})}{4} + \frac{1}{4} (\ln (| - 1 |) - \ln (| - 2 |))$

Step 10.3

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

$- \frac{\ln (\frac{| 3 |}{| 2 |})}{4} + \frac{1}{4} \ln (\frac{| - 1 |}{| - 2 |})$

Step 10.4

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

$- \frac{\ln (\frac{| 3 |}{| 2 |})}{4} + \frac{\ln (\frac{| - 1 |}{| - 2 |})}{4}$

Step 11

Simplify.

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Step 11.1

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$- \frac{\ln (\frac{3}{| 2 |})}{4} + \frac{\ln (\frac{| - 1 |}{| - 2 |})}{4}$

Step 11.2

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$- \frac{\ln (\frac{3}{2})}{4} + \frac{\ln (\frac{| - 1 |}{| - 2 |})}{4}$

Step 11.3

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

$- \frac{\ln (\frac{3}{2})}{4} + \frac{\ln (\frac{1}{| - 2 |})}{4}$

Step 11.4

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

$- \frac{\ln (\frac{3}{2})}{4} + \frac{\ln (\frac{1}{2})}{4}$

Step 12

The result can be shown in multiple forms.

Exact Form:

$- \frac{\ln (\frac{3}{2})}{4} + \frac{\ln (\frac{1}{2})}{4}$

Decimal Form:

$- 0.27465307 \dots$

Step 13