Calculus Examples

$\int \frac{1}{4 - 9 x^{2}} d x d x$

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 the fraction.

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Step 1.1.1.1

Rewrite $4$ as $2^{2}$ .

$\frac{1}{2^{2} - 9 x^{2}}$

Step 1.1.1.2

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

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

Step 1.1.1.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = 3 x$ .

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

Step 1.1.1.4

Multiply $3$ by $- 1$ .

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

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 $A$ .

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

Step 1.1.3

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

Step 1.1.4

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

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

Step 1.1.5

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

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Step 1.1.5.1

Cancel the common factor.

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

Step 1.1.5.2

Rewrite the expression.

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

Step 1.1.6

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

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Step 1.1.6.1

Cancel the common factor.

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

Step 1.1.6.2

Rewrite the expression.

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

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

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

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Step 1.1.7.1.1

Cancel the common factor.

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

Step 1.1.7.1.2

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

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

Step 1.1.7.2

Apply the distributive property.

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

Step 1.1.7.3

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

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

Step 1.1.7.4

Rewrite using the commutative property of multiplication.

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

Step 1.1.7.5

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

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Step 1.1.7.5.1

Cancel the common factor.

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

Step 1.1.7.5.2

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

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

Step 1.1.7.6

Apply the distributive property.

$1 = 2 A - 3 A x + B \cdot 2 + B (3 x)$

Step 1.1.7.7

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

$1 = 2 A - 3 A x + 2 \cdot B + B (3 x)$

Step 1.1.7.8

Rewrite using the commutative property of multiplication.

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

Step 1.1.8

Reorder.

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Step 1.1.8.1

Move $A$ .

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

Step 1.1.8.2

Move $B$ .

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

Step 1.1.8.3

Move $2 B$ .

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

Step 1.1.8.4

Move $2 A$ .

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

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.

$0 = - 3 A + 3 B$

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.

$1 = 2 A + 2 B$

Step 1.2.3

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

$0 = - 3 A + 3 B$

$1 = 2 A + 2 B$

$0 = - 3 A + 3 B$

$1 = 2 A + 2 B$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $A$ in $0 = - 3 A + 3 B$ .

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Step 1.3.1.1

Rewrite the equation as $- 3 A + 3 B = 0$ .

$- 3 A + 3 B = 0$

$1 = 2 A + 2 B$

Step 1.3.1.2

Subtract $3 B$ from both sides of the equation.

$- 3 A = - 3 B$

$1 = 2 A + 2 B$

Step 1.3.1.3

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

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Step 1.3.1.3.1

Divide each term in $- 3 A = - 3 B$ by $- 3$ .

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

$1 = 2 A + 2 B$

Step 1.3.1.3.2

Simplify the left side.

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Step 1.3.1.3.2.1

Cancel the common factor of $- 3$ .

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Step 1.3.1.3.2.1.1

Cancel the common factor.

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

$1 = 2 A + 2 B$

Step 1.3.1.3.2.1.2

Divide $A$ by $1$ .

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

$1 = 2 A + 2 B$

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

$1 = 2 A + 2 B$

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

$1 = 2 A + 2 B$

Step 1.3.1.3.3

Simplify the right side.

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Step 1.3.1.3.3.1

Cancel the common factor of $- 3$ .

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Step 1.3.1.3.3.1.1

Cancel the common factor.

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

$1 = 2 A + 2 B$

Step 1.3.1.3.3.1.2

Divide $B$ by $1$ .

$A = B$

$1 = 2 A + 2 B$

$A = B$

$1 = 2 A + 2 B$

$A = B$

$1 = 2 A + 2 B$

$A = B$

$1 = 2 A + 2 B$

$A = B$

$1 = 2 A + 2 B$

Step 1.3.2

Replace all occurrences of $A$ with $B$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $A$ in $1 = 2 A + 2 B$ with $B$ .

$1 = 2 (B) + 2 B$

$A = B$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Add $2 B$ and $2 B$ .

$1 = 4 B$

$A = B$

$1 = 4 B$

$A = B$

$1 = 4 B$

$A = B$

Step 1.3.3

Solve for $B$ in $1 = 4 B$ .

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Step 1.3.3.1

Rewrite the equation as $4 B = 1$ .

$4 B = 1$

$A = B$

Step 1.3.3.2

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

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Step 1.3.3.2.1

Divide each term in $4 B = 1$ by $4$ .

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

$A = B$

Step 1.3.3.2.2

Simplify the left side.

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Step 1.3.3.2.2.1

Cancel the common factor of $4$ .

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Step 1.3.3.2.2.1.1

Cancel the common factor.

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

$A = B$

Step 1.3.3.2.2.1.2

Divide $B$ by $1$ .

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

$A = B$

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

$A = B$

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

$A = B$

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

$A = B$

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

$A = B$

Step 1.3.4

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

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Step 1.3.4.1

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

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

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

Step 1.3.4.2

Simplify the left side.

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Step 1.3.4.2.1

Remove parentheses.

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

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

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

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

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

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

Step 1.3.5

List all of the solutions.

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

Step 1.4

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

$\frac{\frac{1}{4}}{2 + 3 x} + \frac{\frac{1}{4}}{2 - 3 x}$

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{4} \cdot \frac{1}{2 + 3 x} + \frac{\frac{1}{4}}{2 - 3 x}$

Step 1.5.2

Multiply $\frac{1}{4}$ by $\frac{1}{2 + 3 x}$ .

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

Step 1.5.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.5.4

Multiply $\frac{1}{4}$ by $\frac{1}{2 - 3 x}$ .

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

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

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

Step 4

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

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Step 4.1

Let $u_{1} = 2 + 3 x$ . Find $\frac{d u_{1}}{d x}$ .

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Step 4.1.1

Differentiate $2 + 3 x$ .

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

Step 4.1.2

Differentiate.

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Step 4.1.2.1

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 4.1.2.2

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 4.1.3

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

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Step 4.1.3.1

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 4.1.3.2

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 4.1.3.3

Multiply $3$ by $1$ .

$0 + 3$

Step 4.1.4

Add $0$ and $3$ .

$3$

Step 4.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$(\frac{1}{4} \int \frac{1}{u_{1}} \cdot \frac{1}{3} d u_{1} + \int \frac{1}{4 (2 - 3 x)} d x) d x$

Step 5

Simplify.

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Step 5.1

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

$(\frac{1}{4} \int \frac{1}{u_{1} \cdot 3} d u_{1} + \int \frac{1}{4 (2 - 3 x)} d x) d x$

Step 5.2

Move $3$ to the left of $u_{1}$ .

$(\frac{1}{4} \int \frac{1}{3 u_{1}} d u_{1} + \int \frac{1}{4 (2 - 3 x)} d x) d x$

Step 6

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

$(\frac{1}{4} (\frac{1}{3} \int \frac{1}{u_{1}} d u_{1}) + \int \frac{1}{4 (2 - 3 x)} d x) d x$

Step 7

Simplify.

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Step 7.1

Multiply $\frac{1}{3}$ by $\frac{1}{4}$ .

$(\frac{1}{3 \cdot 4} \int \frac{1}{u_{1}} d u_{1} + \int \frac{1}{4 (2 - 3 x)} d x) d x$

Step 7.2

Multiply $3$ by $4$ .

$(\frac{1}{12} \int \frac{1}{u_{1}} d u_{1} + \int \frac{1}{4 (2 - 3 x)} d x) d x$

Step 8

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

$(\frac{1}{12} (\ln (| u_{1} |) + C) + \int \frac{1}{4 (2 - 3 x)} d x) d x$

Step 9

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

$(\frac{1}{12} (\ln (| u_{1} |) + C) + \frac{1}{4} \int \frac{1}{2 - 3 x} d x) d x$

Step 10

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

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Step 10.1

Let $u_{2} = 2 - 3 x$ . Find $\frac{d u_{2}}{d x}$ .

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Step 10.1.1

Differentiate $2 - 3 x$ .

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

Step 10.1.2

Differentiate.

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Step 10.1.2.1

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 10.1.2.2

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 10.1.3

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

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Step 10.1.3.1

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 10.1.3.2

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 10.1.3.3

Multiply $- 3$ by $1$ .

$0 - 3$

Step 10.1.4

Subtract $3$ from $0$ .

$- 3$

Step 10.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$(\frac{1}{12} (\ln (| u_{1} |) + C) + \frac{1}{4} \int \frac{1}{u_{2}} \cdot \frac{1}{- 3} d u_{2}) d x$

Step 11

Simplify.

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Step 11.1

Move the negative in front of the fraction.

$(\frac{1}{12} (\ln (| u_{1} |) + C) + \frac{1}{4} \int \frac{1}{u_{2}} (- \frac{1}{3}) d u_{2}) d x$

Step 11.2

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

$(\frac{1}{12} (\ln (| u_{1} |) + C) + \frac{1}{4} \int - \frac{1}{u_{2} \cdot 3} d u_{2}) d x$

Step 11.3

Move $3$ to the left of $u_{2}$ .

$(\frac{1}{12} (\ln (| u_{1} |) + C) + \frac{1}{4} \int - \frac{1}{3 u_{2}} d u_{2}) d x$

Step 12

Since $- 1$ is constant with respect to $u_{2}$ , move $- 1$ out of the integral.

$(\frac{1}{12} (\ln (| u_{1} |) + C) + \frac{1}{4} (- \int \frac{1}{3 u_{2}} d u_{2})) d x$

Step 13

Since $\frac{1}{3}$ is constant with respect to $u_{2}$ , move $\frac{1}{3}$ out of the integral.

$(\frac{1}{12} (\ln (| u_{1} |) + C) + \frac{1}{4} (- (\frac{1}{3} \int \frac{1}{u_{2}} d u_{2}))) d x$

Step 14

Simplify.

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Step 14.1

Multiply $\frac{1}{4}$ by $\frac{1}{3}$ .

$(\frac{1}{12} (\ln (| u_{1} |) + C) - \frac{1}{4 \cdot 3} \int \frac{1}{u_{2}} d u_{2}) d x$

Step 14.2

Multiply $4$ by $3$ .

$(\frac{1}{12} (\ln (| u_{1} |) + C) - \frac{1}{12} \int \frac{1}{u_{2}} d u_{2}) d x$

Step 15

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$(\frac{1}{12} (\ln (| u_{1} |) + C) - \frac{1}{12} (\ln (| u_{2} |) + C)) d x$

Step 16

Simplify.

$(\frac{1}{12} \ln (| u_{1} |) - \frac{1}{12} \ln (| u_{2} |)) d x + C$

Step 17

Substitute back in for each integration substitution variable.

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Step 17.1

Replace all occurrences of $u_{1}$ with $2 + 3 x$ .

$(\frac{1}{12} \ln (| 2 + 3 x |) - \frac{1}{12} \ln (| u_{2} |)) d x + C$

Step 17.2

Replace all occurrences of $u_{2}$ with $2 - 3 x$ .

$(\frac{1}{12} \ln (| 2 + 3 x |) - \frac{1}{12} \ln (| 2 - 3 x |)) d x + C$