Calculus Examples

$\int \frac{6}{2 x + 3 x^{2}} d x$

Step 1

Since $6$ is constant with respect to $x$ , move $6$ out of the integral.

$6 \int \frac{1}{2 x + 3 x^{2}} d x$

Step 2

Write the fraction using partial fraction decomposition.

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Step 2.1

Decompose the fraction and multiply through by the common denominator.

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Step 2.1.1

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

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Step 2.1.1.1

Factor $x$ out of $2 x$ .

$\frac{1}{x \cdot 2 + 3 x^{2}}$

Step 2.1.1.2

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

$\frac{1}{x \cdot 2 + x (3 x)}$

Step 2.1.1.3

Factor $x$ out of $x \cdot 2 + x (3 x)$ .

$\frac{1}{x \cdot (2 + 3 x)}$

Step 2.1.1.4

Multiply $x$ by $2 + 3 x$ .

$\frac{1}{x (2 + 3 x)}$

Step 2.1.2

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 $B$ .

$\frac{A}{x} + \frac{B}{2 + 3 x}$

Step 2.1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $x (2 + 3 x)$ .

$\frac{1 (x (2 + 3 x))}{x (2 + 3 x)} = \frac{A (x (2 + 3 x))}{x} + \frac{(B) (x (2 + 3 x))}{2 + 3 x}$

Step 2.1.4

Cancel the common factor of $x$ .

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Step 2.1.4.1

Cancel the common factor.

$\frac{1 (x (2 + 3 x))}{x (2 + 3 x)} = \frac{A (x (2 + 3 x))}{x} + \frac{(B) (x (2 + 3 x))}{2 + 3 x}$

Step 2.1.4.2

Rewrite the expression.

$\frac{1 (2 + 3 x)}{2 + 3 x} = \frac{A (x (2 + 3 x))}{x} + \frac{(B) (x (2 + 3 x))}{2 + 3 x}$

Step 2.1.5

Cancel the common factor of $2 + 3 x$ .

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Step 2.1.5.1

Cancel the common factor.

$\frac{1 (2 + 3 x)}{2 + 3 x} = \frac{A (x (2 + 3 x))}{x} + \frac{(B) (x (2 + 3 x))}{2 + 3 x}$

Step 2.1.5.2

Rewrite the expression.

$1 = \frac{A (x (2 + 3 x))}{x} + \frac{(B) (x (2 + 3 x))}{2 + 3 x}$

Step 2.1.6

Simplify each term.

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Step 2.1.6.1

Cancel the common factor of $x$ .

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Step 2.1.6.1.1

Cancel the common factor.

$1 = \frac{A (x (2 + 3 x))}{x} + \frac{(B) (x (2 + 3 x))}{2 + 3 x}$

Step 2.1.6.1.2

Divide $A (2 + 3 x)$ by $1$ .

$1 = A (2 + 3 x) + \frac{(B) (x (2 + 3 x))}{2 + 3 x}$

Step 2.1.6.2

Apply the distributive property.

$1 = A \cdot 2 + A (3 x) + \frac{(B) (x (2 + 3 x))}{2 + 3 x}$

Step 2.1.6.3

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

$1 = 2 \cdot A + A (3 x) + \frac{(B) (x (2 + 3 x))}{2 + 3 x}$

Step 2.1.6.4

Rewrite using the commutative property of multiplication.

$1 = 2 \cdot A + 3 A x + \frac{(B) (x (2 + 3 x))}{2 + 3 x}$

Step 2.1.6.5

Cancel the common factor of $2 + 3 x$ .

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Step 2.1.6.5.1

Cancel the common factor.

$1 = 2 A + 3 A x + \frac{B (x (2 + 3 x))}{2 + 3 x}$

Step 2.1.6.5.2

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

$1 = 2 A + 3 A x + (B) (x)$

$1 = 2 A + 3 A x + B x$

Step 2.1.7

Simplify the expression.

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Step 2.1.7.1

Move $A$ .

$1 = 2 A + 3 x A + B x$

Step 2.1.7.2

Reorder $B$ and $x$ .

$1 = 2 A + 3 x A + B x$

Step 2.1.7.3

Move $2 A$ .

$1 = 3 x A + B x + 2 A$

Step 2.2

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

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Step 2.2.1

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.

$0 = 3 A + B$

Step 2.2.2

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 2.2.3

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

$0 = 3 A + B$

$1 = 2 A$

$0 = 3 A + B$

$1 = 2 A$

Step 2.3

Solve the system of equations.

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Step 2.3.1

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

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Step 2.3.1.1

Rewrite the equation as $2 A = 1$ .

$2 A = 1$

$0 = 3 A + B$

Step 2.3.1.2

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

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Step 2.3.1.2.1

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

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

$0 = 3 A + B$

Step 2.3.1.2.2

Simplify the left side.

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Step 2.3.1.2.2.1

Cancel the common factor of $2$ .

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Step 2.3.1.2.2.1.1

Cancel the common factor.

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

$0 = 3 A + B$

Step 2.3.1.2.2.1.2

Divide $A$ by $1$ .

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

$0 = 3 A + B$

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

$0 = 3 A + B$

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

$0 = 3 A + B$

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

$0 = 3 A + B$

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

$0 = 3 A + B$

Step 2.3.2

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

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Step 2.3.2.1

Replace all occurrences of $A$ in $0 = 3 A + B$ with $\frac{1}{2}$ .

$0 = 3 (\frac{1}{2}) + B$

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

Step 2.3.2.2

Simplify the right side.

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Step 2.3.2.2.1

Combine $3$ and $\frac{1}{2}$ .

$0 = \frac{3}{2} + B$

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

$0 = \frac{3}{2} + B$

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

$0 = \frac{3}{2} + B$

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

Step 2.3.3

Solve for $B$ in $0 = \frac{3}{2} + B$ .

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Step 2.3.3.1

Rewrite the equation as $\frac{3}{2} + B = 0$ .

$\frac{3}{2} + B = 0$

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

Step 2.3.3.2

Subtract $\frac{3}{2}$ from both sides of the equation.

$B = - \frac{3}{2}$

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

$B = - \frac{3}{2}$

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

Step 2.3.4

Solve the system of equations.

$B = - \frac{3}{2} A = \frac{1}{2}$

Step 2.3.5

List all of the solutions.

$B = - \frac{3}{2}, A = \frac{1}{2}$

Step 2.4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{2 + 3 x}$ with the values found for $A$ and $B$ .

$\frac{\frac{1}{2}}{x} + \frac{- \frac{3}{2}}{2 + 3 x}$

Step 2.5

Simplify.

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Step 2.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{2} \cdot \frac{1}{x} + \frac{- \frac{3}{2}}{2 + 3 x}$

Step 2.5.2

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

$\frac{1}{2 x} + \frac{- \frac{3}{2}}{2 + 3 x}$

Step 2.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{2 x} - \frac{3}{2} \cdot \frac{1}{2 + 3 x}$

Step 2.5.4

Multiply $\frac{1}{2 + 3 x}$ by $\frac{3}{2}$ .

$\frac{1}{2 x} - \frac{3}{(2 + 3 x) \cdot 2}$

Step 2.5.5

Move $2$ to the left of $2 + 3 x$ .

$6 \int \frac{1}{2 x} - \frac{3}{2 (2 + 3 x)} d x$

Step 3

Split the single integral into multiple integrals.

$6 (\int \frac{1}{2 x} d x + \int - \frac{3}{2 (2 + 3 x)} d x)$

Step 4

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

$6 (\frac{1}{2} \int \frac{1}{x} d x + \int - \frac{3}{2 (2 + 3 x)} d x)$

Step 5

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

$6 (\frac{1}{2} (\ln (| x |) + C) + \int - \frac{3}{2 (2 + 3 x)} d x)$

Step 6

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

$6 (\frac{1}{2} (\ln (| x |) + C) - \int \frac{3}{2 (2 + 3 x)} d x)$

Step 7

Since $\frac{3}{2}$ is constant with respect to $x$ , move $\frac{3}{2}$ out of the integral.

$6 (\frac{1}{2} (\ln (| x |) + C) - (\frac{3}{2} \int \frac{1}{2 + 3 x} d x))$

Step 8

Let $u = 2 + 3 x$ . Then $d u = 3 d x$ , so $\frac{1}{3} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 8.1

Let $u = 2 + 3 x$ . Find $\frac{d u}{d x}$ .

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Step 8.1.1

Differentiate $2 + 3 x$ .

$\frac{d}{d x} [2 + 3 x]$

Step 8.1.2

Differentiate.

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Step 8.1.2.1

By the Sum Rule, the derivative of $2 + 3 x$ with respect to $x$ is $\frac{d}{d x} [2] + \frac{d}{d x} [3 x]$ .

$\frac{d}{d x} [2] + \frac{d}{d x} [3 x]$

Step 8.1.2.2

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

$0 + \frac{d}{d x} [3 x]$

Step 8.1.3

Evaluate $\frac{d}{d x} [3 x]$ .

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Step 8.1.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$0 + 3 \frac{d}{d x} [x]$

Step 8.1.3.2

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

$0 + 3 \cdot 1$

Step 8.1.3.3

Multiply $3$ by $1$ .

$0 + 3$

Step 8.1.4

Add $0$ and $3$ .

$3$

Step 8.2

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

$6 (\frac{1}{2} (\ln (| x |) + C) - \frac{3}{2} \int \frac{1}{u} \cdot \frac{1}{3} d u)$

Step 9

Simplify.

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Step 9.1

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

$6 (\frac{1}{2} (\ln (| x |) + C) - \frac{3}{2} \int \frac{1}{u \cdot 3} d u)$

Step 9.2

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

$6 (\frac{1}{2} (\ln (| x |) + C) - \frac{3}{2} \int \frac{1}{3 u} d u)$

Step 10

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

$6 (\frac{1}{2} (\ln (| x |) + C) - \frac{3}{2} (\frac{1}{3} \int \frac{1}{u} d u))$

Step 11

Simplify.

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Step 11.1

Multiply $\frac{1}{3}$ by $\frac{3}{2}$ .

$6 (\frac{1}{2} (\ln (| x |) + C) - \frac{3}{3 \cdot 2} \int \frac{1}{u} d u)$

Step 11.2

Multiply $3$ by $2$ .

$6 (\frac{1}{2} (\ln (| x |) + C) - \frac{3}{6} \int \frac{1}{u} d u)$

Step 11.3

Cancel the common factor of $3$ and $6$ .

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Step 11.3.1

Factor $3$ out of $3$ .

$6 (\frac{1}{2} (\ln (| x |) + C) - \frac{3 (1)}{6} \int \frac{1}{u} d u)$

Step 11.3.2

Cancel the common factors.

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Step 11.3.2.1

Factor $3$ out of $6$ .

$6 (\frac{1}{2} (\ln (| x |) + C) - \frac{3 \cdot 1}{3 \cdot 2} \int \frac{1}{u} d u)$

Step 11.3.2.2

Cancel the common factor.

$6 (\frac{1}{2} (\ln (| x |) + C) - \frac{3 \cdot 1}{3 \cdot 2} \int \frac{1}{u} d u)$

Step 11.3.2.3

Rewrite the expression.

$6 (\frac{1}{2} (\ln (| x |) + C) - \frac{1}{2} \int \frac{1}{u} d u)$

Step 12

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

$6 (\frac{1}{2} (\ln (| x |) + C) - \frac{1}{2} (\ln (| u |) + C))$

Step 13

Simplify.

$6 (\frac{1}{2} \ln (| x |) - \frac{1}{2} \ln (| u |)) + C$

Step 14

Replace all occurrences of $u$ with $2 + 3 x$ .

$6 (\frac{1}{2} \ln (| x |) - \frac{1}{2} \ln (| 2 + 3 x |)) + C$

Step 15

Simplify.

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Step 15.1

Simplify each term.

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Step 15.1.1

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

$6 (\frac{\ln (| x |)}{2} - \frac{1}{2} \ln (| 2 + 3 x |)) + C$

Step 15.1.2

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

$6 (\frac{\ln (| x |)}{2} - \frac{\ln (| 2 + 3 x |)}{2}) + C$

Step 15.2

Combine the numerators over the common denominator.

$6 \frac{\ln (| x |) - \ln (| 2 + 3 x |)}{2} + C$

Step 15.3

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

$6 \frac{\ln (\frac{| x |}{| 2 + 3 x |})}{2} + C$

Step 15.4

Cancel the common factor of $2$ .

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Step 15.4.1

Factor $2$ out of $6$ .

$2 (3) \frac{\ln (\frac{| x |}{| 2 + 3 x |})}{2} + C$

Step 15.4.2

Cancel the common factor.

$2 \cdot 3 \frac{\ln (\frac{| x |}{| 2 + 3 x |})}{2} + C$

Step 15.4.3

Rewrite the expression.

$3 \ln (\frac{| x |}{| 2 + 3 x |}) + C$