Calculus Examples

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

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 $2^{2}$ .

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

Step 1.1.1.2

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 $b = x$ .

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

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 $A$ .

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

Step 1.1.3

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

Step 1.1.4

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

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

Step 1.1.5

Cancel the common factor of $2 + x$ .

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Step 1.1.5.1

Cancel the common factor.

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

Step 1.1.5.2

Rewrite the expression.

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

Step 1.1.6

Cancel the common factor of $2 - x$ .

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Step 1.1.6.1

Cancel the common factor.

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

Step 1.1.6.2

Rewrite the expression.

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

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

Cancel the common factor of $2 + x$ .

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Step 1.1.7.1.1

Cancel the common factor.

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

Step 1.1.7.1.2

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

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

Step 1.1.7.2

Apply the distributive property.

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

Step 1.1.7.3

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

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

Step 1.1.7.4

Rewrite using the commutative property of multiplication.

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

Step 1.1.7.5

Cancel the common factor of $2 - x$ .

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Step 1.1.7.5.1

Cancel the common factor.

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

Step 1.1.7.5.2

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

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

Step 1.1.7.6

Apply the distributive property.

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

Step 1.1.7.7

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

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

Step 1.1.8

Simplify the expression.

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Step 1.1.8.1

Move $A$ .

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

Step 1.1.8.2

Reorder $B$ and $x$ .

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

Step 1.1.8.3

Move $2 B$ .

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

Step 1.1.8.4

Move $2 A$ .

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

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.

$0 = - 1 A + B$

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.

$1 = 2 A + 2 B$

Step 1.2.3

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

$0 = - 1 A + B$

$1 = 2 A + 2 B$

$0 = - 1 A + B$

$1 = 2 A + 2 B$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $B$ in $0 = - 1 A + B$ .

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Step 1.3.1.1

Rewrite the equation as $- 1 A + B = 0$ .

$- 1 A + B = 0$

$1 = 2 A + 2 B$

Step 1.3.1.2

Rewrite $- 1 A$ as $- A$ .

$- A + B = 0$

$1 = 2 A + 2 B$

Step 1.3.1.3

Add $A$ to both sides of the equation.

$B = A$

$1 = 2 A + 2 B$

$B = A$

$1 = 2 A + 2 B$

Step 1.3.2

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

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Step 1.3.2.1

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

$1 = 2 A + 2 (A)$

$B = A$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Add $2 A$ and $2 A$ .

$1 = 4 A$

$B = A$

$1 = 4 A$

$B = A$

$1 = 4 A$

$B = A$

Step 1.3.3

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

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Step 1.3.3.1

Rewrite the equation as $4 A = 1$ .

$4 A = 1$

$B = A$

Step 1.3.3.2

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

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Step 1.3.3.2.1

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

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

$B = A$

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.

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

$B = A$

Step 1.3.3.2.2.1.2

Divide $A$ by $1$ .

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

$B = A$

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

$B = A$

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

$B = A$

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

$B = A$

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

$B = A$

Step 1.3.4

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

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Step 1.3.4.1

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

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

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

Step 1.3.4.2

Simplify the left side.

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Step 1.3.4.2.1

Remove parentheses.

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

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

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

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

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

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

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 $\frac{A}{2 + x} + \frac{B}{2 - x}$ with the values found for $A$ and $B$ .

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

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.5.2

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

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

Step 1.5.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.5.4

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

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

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

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

Step 4

Let $u_{1} = 2 + x$ . Then $d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 4.1

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

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Step 4.1.1

Differentiate $2 + x$ .

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

Step 4.1.2

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

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

Step 4.1.3

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

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

Step 4.1.4

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

$0 + 1$

Step 4.1.5

Add $0$ and $1$ .

$1$

Step 4.2

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

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

Step 5

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

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

Step 6

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

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

Step 7

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

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Step 7.1

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

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Step 7.1.1

Rewrite.

$\frac{- 1}{1}$

Step 7.1.2

Divide $- 1$ by $1$ .

$- 1$

Step 7.2

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

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

Step 8

Move the negative in front of the fraction.

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

Step 9

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

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

Step 10

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

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

Step 11

Simplify.

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Step 11.1

Simplify.

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

Step 11.2

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

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

Step 12

Substitute back in for each integration substitution variable.

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Step 12.1

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

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

Step 12.2

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

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

Step 13

Reorder terms.

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