Calculus Examples

$\int \frac{1}{25 - 4 x^{2}} d x$

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 $25$ as $5^{2}$ .

$\frac{1}{5^{2} - 4 x^{2}}$

Step 1.1.1.2

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$\frac{1}{5^{2} - {(2 x)}^{2}}$

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 $a = 5$ and $b = 2 x$ .

$\frac{1}{(5 + 2 x) (5 - (2 x))}$

Step 1.1.1.4

Multiply $2$ by $- 1$ .

$\frac{1}{(5 + 2 x) (5 - 2 x)}$

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 $A$ .

$\frac{A}{5 + 2 x}$

Step 1.1.3

$\frac{A}{5 + 2 x} + \frac{B}{5 - 2 x}$

Step 1.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(5 + 2 x) (5 - 2 x)$ .

$\frac{1 (5 + 2 x) (5 - 2 x)}{(5 + 2 x) (5 - 2 x)} = \frac{(A) (5 + 2 x) (5 - 2 x)}{5 + 2 x} + \frac{(B) (5 + 2 x) (5 - 2 x)}{5 - 2 x}$

Step 1.1.5

Cancel the common factor of $5 + 2 x$ .

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Step 1.1.5.1

Cancel the common factor.

$\frac{1 (5 + 2 x) (5 - 2 x)}{(5 + 2 x) (5 - 2 x)} = \frac{(A) (5 + 2 x) (5 - 2 x)}{5 + 2 x} + \frac{(B) (5 + 2 x) (5 - 2 x)}{5 - 2 x}$

Step 1.1.5.2

Rewrite the expression.

$\frac{1 (5 - 2 x)}{5 - 2 x} = \frac{(A) (5 + 2 x) (5 - 2 x)}{5 + 2 x} + \frac{(B) (5 + 2 x) (5 - 2 x)}{5 - 2 x}$

Step 1.1.6

Cancel the common factor of $5 - 2 x$ .

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Step 1.1.6.1

Cancel the common factor.

$\frac{1 (5 - 2 x)}{5 - 2 x} = \frac{(A) (5 + 2 x) (5 - 2 x)}{5 + 2 x} + \frac{(B) (5 + 2 x) (5 - 2 x)}{5 - 2 x}$

Step 1.1.6.2

Rewrite the expression.

$1 = \frac{(A) (5 + 2 x) (5 - 2 x)}{5 + 2 x} + \frac{(B) (5 + 2 x) (5 - 2 x)}{5 - 2 x}$

Step 1.1.7

Simplify each term.

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Step 1.1.7.1

Cancel the common factor of $5 + 2 x$ .

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Step 1.1.7.1.1

Cancel the common factor.

$1 = \frac{A (5 + 2 x) (5 - 2 x)}{5 + 2 x} + \frac{(B) (5 + 2 x) (5 - 2 x)}{5 - 2 x}$

Step 1.1.7.1.2

Divide $(A) (5 - 2 x)$ by $1$ .

$1 = (A) (5 - 2 x) + \frac{(B) (5 + 2 x) (5 - 2 x)}{5 - 2 x}$

Step 1.1.7.2

Apply the distributive property.

$1 = A \cdot 5 + A (- 2 x) + \frac{(B) (5 + 2 x) (5 - 2 x)}{5 - 2 x}$

Step 1.1.7.3

Move $5$ to the left of $A$ .

$1 = 5 \cdot A + A (- 2 x) + \frac{(B) (5 + 2 x) (5 - 2 x)}{5 - 2 x}$

Step 1.1.7.4

Rewrite using the commutative property of multiplication.

$1 = 5 \cdot A - 2 A x + \frac{(B) (5 + 2 x) (5 - 2 x)}{5 - 2 x}$

Step 1.1.7.5

Cancel the common factor of $5 - 2 x$ .

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Step 1.1.7.5.1

Cancel the common factor.

$1 = 5 A - 2 A x + \frac{(B) (5 + 2 x) (5 - 2 x)}{5 - 2 x}$

Step 1.1.7.5.2

Divide $(B) (5 + 2 x)$ by $1$ .

$1 = 5 A - 2 A x + (B) (5 + 2 x)$

Step 1.1.7.6

Apply the distributive property.

$1 = 5 A - 2 A x + B \cdot 5 + B (2 x)$

Step 1.1.7.7

Move $5$ to the left of $B$ .

$1 = 5 A - 2 A x + 5 \cdot B + B (2 x)$

Step 1.1.7.8

Rewrite using the commutative property of multiplication.

$1 = 5 A - 2 A x + 5 B + 2 B x$

Step 1.1.8

Reorder.

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Step 1.1.8.1

Move $A$ .

$1 = 5 A - 2 x A + 5 B + 2 B x$

Step 1.1.8.2

Move $B$ .

$1 = 5 A - 2 x A + 5 B + 2 x B$

Step 1.1.8.3

Move $5 B$ .

$1 = 5 A - 2 x A + 2 x B + 5 B$

Step 1.1.8.4

Move $5 A$ .

$1 = - 2 x A + 2 x B + 5 A + 5 B$

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.

$0 = - 2 A + 2 B$

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.

$1 = 5 A + 5 B$

Step 1.2.3

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

$0 = - 2 A + 2 B$

$1 = 5 A + 5 B$

$0 = - 2 A + 2 B$

$1 = 5 A + 5 B$

Step 1.3

Solve the system of equations.

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Step 1.3.1

Solve for $A$ in $0 = - 2 A + 2 B$ .

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Step 1.3.1.1

Rewrite the equation as $- 2 A + 2 B = 0$ .

$- 2 A + 2 B = 0$

$1 = 5 A + 5 B$

Step 1.3.1.2

Subtract $2 B$ from both sides of the equation.

$- 2 A = - 2 B$

$1 = 5 A + 5 B$

Step 1.3.1.3

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

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Step 1.3.1.3.1

Divide each term in $- 2 A = - 2 B$ by $- 2$ .

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

$1 = 5 A + 5 B$

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 $- 2$ .

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Step 1.3.1.3.2.1.1

Cancel the common factor.

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

$1 = 5 A + 5 B$

Step 1.3.1.3.2.1.2

Divide $A$ by $1$ .

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

$1 = 5 A + 5 B$

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

$1 = 5 A + 5 B$

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

$1 = 5 A + 5 B$

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 $- 2$ .

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Step 1.3.1.3.3.1.1

Cancel the common factor.

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

$1 = 5 A + 5 B$

Step 1.3.1.3.3.1.2

Divide $B$ by $1$ .

$A = B$

$1 = 5 A + 5 B$

$A = B$

$1 = 5 A + 5 B$

$A = B$

$1 = 5 A + 5 B$

$A = B$

$1 = 5 A + 5 B$

$A = B$

$1 = 5 A + 5 B$

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 $1 = 5 A + 5 B$ with $B$ .

$1 = 5 (B) + 5 B$

$A = B$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Add $5 B$ and $5 B$ .

$1 = 10 B$

$A = B$

$1 = 10 B$

$A = B$

$1 = 10 B$

$A = B$

Step 1.3.3

Solve for $B$ in $1 = 10 B$ .

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Step 1.3.3.1

Rewrite the equation as $10 B = 1$ .

$10 B = 1$

$A = B$

Step 1.3.3.2

Divide each term in $10 B = 1$ by $10$ and simplify.

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Step 1.3.3.2.1

Divide each term in $10 B = 1$ by $10$ .

$\frac{10 B}{10} = \frac{1}{10}$

$A = B$

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 $10$ .

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Step 1.3.3.2.2.1.1

Cancel the common factor.

$\frac{10 B}{10} = \frac{1}{10}$

$A = B$

Step 1.3.3.2.2.1.2

Divide $B$ by $1$ .

$B = \frac{1}{10}$

$A = B$

$B = \frac{1}{10}$

$A = B$

$B = \frac{1}{10}$

$A = B$

$B = \frac{1}{10}$

$A = B$

$B = \frac{1}{10}$

$A = B$

Step 1.3.4

Replace all occurrences of $B$ with $\frac{1}{10}$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $B$ in $A = B$ with $\frac{1}{10}$ .

$A = \frac{1}{10}$

$B = \frac{1}{10}$

Step 1.3.4.2

Simplify the left side.

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Step 1.3.4.2.1

Remove parentheses.

$A = \frac{1}{10}$

$B = \frac{1}{10}$

$A = \frac{1}{10}$

$B = \frac{1}{10}$

$A = \frac{1}{10}$

$B = \frac{1}{10}$

Step 1.3.5

List all of the solutions.

$A = \frac{1}{10}, B = \frac{1}{10}$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{5 + 2 x} + \frac{B}{5 - 2 x}$ with the values found for $A$ and $B$ .

$\frac{\frac{1}{10}}{5 + 2 x} + \frac{\frac{1}{10}}{5 - 2 x}$

Step 1.5

Simplify.

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Step 1.5.1

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{10} \cdot \frac{1}{5 + 2 x} + \frac{\frac{1}{10}}{5 - 2 x}$

Step 1.5.2

Multiply $\frac{1}{10}$ by $\frac{1}{5 + 2 x}$ .

$\frac{1}{10 (5 + 2 x)} + \frac{\frac{1}{10}}{5 - 2 x}$

Step 1.5.3

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{10 (5 + 2 x)} + \frac{1}{10} \cdot \frac{1}{5 - 2 x}$

Step 1.5.4

Multiply $\frac{1}{10}$ by $\frac{1}{5 - 2 x}$ .

$\int \frac{1}{10 (5 + 2 x)} + \frac{1}{10 (5 - 2 x)} d x$

Step 2

Split the single integral into multiple integrals.

$\int \frac{1}{10 (5 + 2 x)} d x + \int \frac{1}{10 (5 - 2 x)} d x$

Step 3

Since $\frac{1}{10}$ is constant with respect to $x$ , move $\frac{1}{10}$ out of the integral.

$\frac{1}{10} \int \frac{1}{5 + 2 x} d x + \int \frac{1}{10 (5 - 2 x)} d x$

Step 4

Let $u_{1} = 5 + 2 x$ . Then $d u_{1} = 2 d x$ , so $\frac{1}{2} d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 4.1

Let $u_{1} = 5 + 2 x$ . Find $\frac{d u_{1}}{d x}$ .

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Step 4.1.1

Differentiate $5 + 2 x$ .

$\frac{d}{d x} [5 + 2 x]$

Step 4.1.2

Differentiate.

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Step 4.1.2.1

By the Sum Rule, the derivative of $5 + 2 x$ with respect to $x$ is $\frac{d}{d x} [5] + \frac{d}{d x} [2 x]$ .

$\frac{d}{d x} [5] + \frac{d}{d x} [2 x]$

Step 4.1.2.2

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [2 x]$

Step 4.1.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 4.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$0 + 2 \frac{d}{d x} [x]$

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 $n = 1$ .

$0 + 2 \cdot 1$

Step 4.1.3.3

Multiply $2$ by $1$ .

$0 + 2$

Step 4.1.4

Add $0$ and $2$ .

$2$

Step 4.2

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

$\frac{1}{10} \int \frac{1}{u_{1}} \cdot \frac{1}{2} d u_{1} + \int \frac{1}{10 (5 - 2 x)} d x$

Step 5

Simplify.

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Step 5.1

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

$\frac{1}{10} \int \frac{1}{u_{1} \cdot 2} d u_{1} + \int \frac{1}{10 (5 - 2 x)} d x$

Step 5.2

Move $2$ to the left of $u_{1}$ .

$\frac{1}{10} \int \frac{1}{2 u_{1}} d u_{1} + \int \frac{1}{10 (5 - 2 x)} d x$

Step 6

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

$\frac{1}{10} (\frac{1}{2} \int \frac{1}{u_{1}} d u_{1}) + \int \frac{1}{10 (5 - 2 x)} d x$

Step 7

Simplify.

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Step 7.1

Multiply $\frac{1}{2}$ by $\frac{1}{10}$ .

$\frac{1}{2 \cdot 10} \int \frac{1}{u_{1}} d u_{1} + \int \frac{1}{10 (5 - 2 x)} d x$

Step 7.2

Multiply $2$ by $10$ .

$\frac{1}{20} \int \frac{1}{u_{1}} d u_{1} + \int \frac{1}{10 (5 - 2 x)} d x$

Step 8

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

$\frac{1}{20} (\ln (| u_{1} |) + C) + \int \frac{1}{10 (5 - 2 x)} d x$

Step 9

Since $\frac{1}{10}$ is constant with respect to $x$ , move $\frac{1}{10}$ out of the integral.

$\frac{1}{20} (\ln (| u_{1} |) + C) + \frac{1}{10} \int \frac{1}{5 - 2 x} d x$

Step 10

Let $u_{2} = 5 - 2 x$ . Then $d u_{2} = - 2 d x$ , so $- \frac{1}{2} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 10.1

Let $u_{2} = 5 - 2 x$ . Find $\frac{d u_{2}}{d x}$ .

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Step 10.1.1

Differentiate $5 - 2 x$ .

$\frac{d}{d x} [5 - 2 x]$

Step 10.1.2

Differentiate.

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Step 10.1.2.1

By the Sum Rule, the derivative of $5 - 2 x$ with respect to $x$ is $\frac{d}{d x} [5] + \frac{d}{d x} [- 2 x]$ .

$\frac{d}{d x} [5] + \frac{d}{d x} [- 2 x]$

Step 10.1.2.2

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- 2 x]$

Step 10.1.3

Evaluate $\frac{d}{d x} [- 2 x]$ .

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Step 10.1.3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

$0 - 2 \frac{d}{d x} [x]$

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 $n = 1$ .

$0 - 2 \cdot 1$

Step 10.1.3.3

Multiply $- 2$ by $1$ .

$0 - 2$

Step 10.1.4

Subtract $2$ from $0$ .

$- 2$

Step 10.2

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

$\frac{1}{20} (\ln (| u_{1} |) + C) + \frac{1}{10} \int \frac{1}{u_{2}} \cdot \frac{1}{- 2} d u_{2}$

Step 11

Simplify.

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Step 11.1

Move the negative in front of the fraction.

$\frac{1}{20} (\ln (| u_{1} |) + C) + \frac{1}{10} \int \frac{1}{u_{2}} (- \frac{1}{2}) d u_{2}$

Step 11.2

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

$\frac{1}{20} (\ln (| u_{1} |) + C) + \frac{1}{10} \int - \frac{1}{u_{2} \cdot 2} d u_{2}$

Step 11.3

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

$\frac{1}{20} (\ln (| u_{1} |) + C) + \frac{1}{10} \int - \frac{1}{2 u_{2}} d u_{2}$

Step 12

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

$\frac{1}{20} (\ln (| u_{1} |) + C) + \frac{1}{10} (- \int \frac{1}{2 u_{2}} d u_{2})$

Step 13

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

$\frac{1}{20} (\ln (| u_{1} |) + C) + \frac{1}{10} (- (\frac{1}{2} \int \frac{1}{u_{2}} d u_{2}))$

Step 14

Simplify.

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Step 14.1

Multiply $\frac{1}{10}$ by $\frac{1}{2}$ .

$\frac{1}{20} (\ln (| u_{1} |) + C) - \frac{1}{10 \cdot 2} \int \frac{1}{u_{2}} d u_{2}$

Step 14.2

Multiply $10$ by $2$ .

$\frac{1}{20} (\ln (| u_{1} |) + C) - \frac{1}{20} \int \frac{1}{u_{2}} d u_{2}$

Step 15

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

$\frac{1}{20} (\ln (| u_{1} |) + C) - \frac{1}{20} (\ln (| u_{2} |) + C)$

Step 16

Simplify.

$\frac{1}{20} \ln (| u_{1} |) - \frac{1}{20} \ln (| u_{2} |) + C$

Step 17

Substitute back in for each integration substitution variable.

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Step 17.1

Replace all occurrences of $u_{1}$ with $5 + 2 x$ .

$\frac{1}{20} \ln (| 5 + 2 x |) - \frac{1}{20} \ln (| u_{2} |) + C$

Step 17.2

Replace all occurrences of $u_{2}$ with $5 - 2 x$ .

$\frac{1}{20} \ln (| 5 + 2 x |) - \frac{1}{20} \ln (| 5 - 2 x |) + C$