Calculus Examples

$\frac{3}{x^{2} + 3 x}$

Step 1

Write $\frac{3}{x^{2} + 3 x}$ as a function.

$f (x) = \frac{3}{x^{2} + 3 x}$

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.

$F (x) = \int \frac{3}{x^{2} + 3 x} d x$

Step 4

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$3 \int \frac{1}{x^{2} + 3 x} d x$

Step 5

Write the fraction using partial fraction decomposition.

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Step 5.1

Decompose the fraction and multiply through by the common denominator.

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Step 5.1.1

Factor $x$ out of $x^{2} + 3 x$ .

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Step 5.1.1.1

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

$\frac{1}{x \cdot x + 3 x}$

Step 5.1.1.2

Factor $x$ out of $3 x$ .

$\frac{1}{x \cdot x + x \cdot 3}$

Step 5.1.1.3

Factor $x$ out of $x \cdot x + x \cdot 3$ .

$\frac{1}{x (x + 3)}$

Step 5.1.2

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 $B$ .

$\frac{A}{x} + \frac{B}{x + 3}$

Step 5.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $x (x + 3)$ .

$\frac{1 (x (x + 3))}{x (x + 3)} = \frac{A (x (x + 3))}{x} + \frac{(B) (x (x + 3))}{x + 3}$

Step 5.1.4

Cancel the common factor of $x$ .

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Step 5.1.4.1

Cancel the common factor.

$\frac{1 (x (x + 3))}{x (x + 3)} = \frac{A (x (x + 3))}{x} + \frac{(B) (x (x + 3))}{x + 3}$

Step 5.1.4.2

Rewrite the expression.

$\frac{1 (x + 3)}{x + 3} = \frac{A (x (x + 3))}{x} + \frac{(B) (x (x + 3))}{x + 3}$

Step 5.1.5

Cancel the common factor of $x + 3$ .

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Step 5.1.5.1

Cancel the common factor.

$\frac{1 (x + 3)}{x + 3} = \frac{A (x (x + 3))}{x} + \frac{(B) (x (x + 3))}{x + 3}$

Step 5.1.5.2

Rewrite the expression.

$1 = \frac{A (x (x + 3))}{x} + \frac{(B) (x (x + 3))}{x + 3}$

Step 5.1.6

Simplify each term.

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Step 5.1.6.1

Cancel the common factor of $x$ .

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Step 5.1.6.1.1

Cancel the common factor.

$1 = \frac{A (x (x + 3))}{x} + \frac{(B) (x (x + 3))}{x + 3}$

Step 5.1.6.1.2

Divide $A (x + 3)$ by $1$ .

$1 = A (x + 3) + \frac{(B) (x (x + 3))}{x + 3}$

Step 5.1.6.2

Apply the distributive property.

$1 = A x + A \cdot 3 + \frac{(B) (x (x + 3))}{x + 3}$

Step 5.1.6.3

Move $3$ to the left of $A$ .

$1 = A x + 3 \cdot A + \frac{(B) (x (x + 3))}{x + 3}$

Step 5.1.6.4

Cancel the common factor of $x + 3$ .

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Step 5.1.6.4.1

Cancel the common factor.

$1 = A x + 3 A + \frac{B (x (x + 3))}{x + 3}$

Step 5.1.6.4.2

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

$1 = A x + 3 A + (B) (x)$

$1 = A x + 3 A + B x$

Step 5.1.7

Move $3 A$ .

$1 = A x + B x + 3 A$

Step 5.2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 5.2.1

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 5.2.2

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.

$1 = 3 A$

Step 5.2.3

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

$0 = A + B$

$1 = 3 A$

$0 = A + B$

$1 = 3 A$

Step 5.3

Solve the system of equations.

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Step 5.3.1

Solve for $A$ in $1 = 3 A$ .

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Step 5.3.1.1

Rewrite the equation as $3 A = 1$ .

$3 A = 1$

$0 = A + B$

Step 5.3.1.2

Divide each term in $3 A = 1$ by $3$ and simplify.

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Step 5.3.1.2.1

Divide each term in $3 A = 1$ by $3$ .

$\frac{3 A}{3} = \frac{1}{3}$

$0 = A + B$

Step 5.3.1.2.2

Simplify the left side.

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Step 5.3.1.2.2.1

Cancel the common factor of $3$ .

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Step 5.3.1.2.2.1.1

Cancel the common factor.

$\frac{3 A}{3} = \frac{1}{3}$

$0 = A + B$

Step 5.3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{1}{3}$

$0 = A + B$

$A = \frac{1}{3}$

$0 = A + B$

$A = \frac{1}{3}$

$0 = A + B$

$A = \frac{1}{3}$

$0 = A + B$

$A = \frac{1}{3}$

$0 = A + B$

Step 5.3.2

Replace all occurrences of $A$ with $\frac{1}{3}$ in each equation.

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Step 5.3.2.1

Replace all occurrences of $A$ in $0 = A + B$ with $\frac{1}{3}$ .

$0 = (\frac{1}{3}) + B$

$A = \frac{1}{3}$

Step 5.3.2.2

Simplify the right side.

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Step 5.3.2.2.1

Remove parentheses.

$0 = \frac{1}{3} + B$

$A = \frac{1}{3}$

$0 = \frac{1}{3} + B$

$A = \frac{1}{3}$

$0 = \frac{1}{3} + B$

$A = \frac{1}{3}$

Step 5.3.3

Solve for $B$ in $0 = \frac{1}{3} + B$ .

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Step 5.3.3.1

Rewrite the equation as $\frac{1}{3} + B = 0$ .

$\frac{1}{3} + B = 0$

$A = \frac{1}{3}$

Step 5.3.3.2

Subtract $\frac{1}{3}$ from both sides of the equation.

$B = - \frac{1}{3}$

$A = \frac{1}{3}$

$B = - \frac{1}{3}$

$A = \frac{1}{3}$

Step 5.3.4

Solve the system of equations.

$B = - \frac{1}{3} A = \frac{1}{3}$

Step 5.3.5

List all of the solutions.

$B = - \frac{1}{3}, A = \frac{1}{3}$

Step 5.4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{x + 3}$ with the values found for $A$ and $B$ .

$\frac{\frac{1}{3}}{x} + \frac{- \frac{1}{3}}{x + 3}$

Step 5.5

Simplify.

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Step 5.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{3} \cdot \frac{1}{x} + \frac{- \frac{1}{3}}{x + 3}$

Step 5.5.2

Multiply $\frac{1}{3}$ by $\frac{1}{x}$ .

$\frac{1}{3 x} + \frac{- \frac{1}{3}}{x + 3}$

Step 5.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{3 x} - \frac{1}{3} \cdot \frac{1}{x + 3}$

Step 5.5.4

Multiply $\frac{1}{x + 3}$ by $\frac{1}{3}$ .

$\frac{1}{3 x} - \frac{1}{(x + 3) \cdot 3}$

Step 5.5.5

Move $3$ to the left of $x + 3$ .

$3 \int \frac{1}{3 x} - \frac{1}{3 (x + 3)} d x$

Step 6

Split the single integral into multiple integrals.

$3 (\int \frac{1}{3 x} d x + \int - \frac{1}{3 (x + 3)} d x)$

Step 7

Since $\frac{1}{3}$ is constant with respect to $x$ , move $\frac{1}{3}$ out of the integral.

$3 (\frac{1}{3} \int \frac{1}{x} d x + \int - \frac{1}{3 (x + 3)} d x)$

Step 8

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

$3 (\frac{1}{3} (\ln (| x |) + C) + \int - \frac{1}{3 (x + 3)} d x)$

Step 9

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

$3 (\frac{1}{3} (\ln (| x |) + C) - \int \frac{1}{3 (x + 3)} d x)$

Step 10

Since $\frac{1}{3}$ is constant with respect to $x$ , move $\frac{1}{3}$ out of the integral.

$3 (\frac{1}{3} (\ln (| x |) + C) - (\frac{1}{3} \int \frac{1}{x + 3} d x))$

Step 11

Let $u = x + 3$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 11.1

Let $u = x + 3$ . Find $\frac{d u}{d x}$ .

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Step 11.1.1

Differentiate $x + 3$ .

$\frac{d}{d x} [x + 3]$

Step 11.1.2

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [3]$

Step 11.1.3

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

$1 + \frac{d}{d x} [3]$

Step 11.1.4

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$1 + 0$

Step 11.1.5

Add $1$ and $0$ .

$1$

Step 11.2

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

$3 (\frac{1}{3} (\ln (| x |) + C) - \frac{1}{3} \int \frac{1}{u} d u)$

Step 12

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

$3 (\frac{1}{3} (\ln (| x |) + C) - \frac{1}{3} (\ln (| u |) + C))$

Step 13

Simplify.

$3 (\frac{1}{3} \ln (| x |) - \frac{1}{3} \ln (| u |)) + C$

Step 14

Replace all occurrences of $u$ with $x + 3$ .

$3 (\frac{1}{3} \ln (| x |) - \frac{1}{3} \ln (| x + 3 |)) + C$

Step 15

Simplify.

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Step 15.1

Simplify each term.

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Step 15.1.1

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

$3 (\frac{\ln (| x |)}{3} - \frac{1}{3} \ln (| x + 3 |)) + C$

Step 15.1.2

Combine $\ln (| x + 3 |)$ and $\frac{1}{3}$ .

$3 (\frac{\ln (| x |)}{3} - \frac{\ln (| x + 3 |)}{3}) + C$

Step 15.2

Combine the numerators over the common denominator.

$3 \frac{\ln (| x |) - \ln (| x + 3 |)}{3} + C$

Step 15.3

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

$3 \frac{\ln (\frac{| x |}{| x + 3 |})}{3} + C$

Step 15.4

Cancel the common factor of $3$ .

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Step 15.4.1

Cancel the common factor.

$3 \frac{\ln (\frac{| x |}{| x + 3 |})}{3} + C$

Step 15.4.2

Rewrite the expression.

$\ln (\frac{| x |}{| x + 3 |}) + C$

Step 16

The answer is the antiderivative of the function $f (x) = \frac{3}{x^{2} + 3 x}$ .

$F (x) =$ $\ln (\frac{| x |}{| x + 3 |}) + C$