Calculus Examples

${(e^{x} + e^{- x})}^{2}$

Step 1

Write ${(e^{x} + e^{- x})}^{2}$ as a function.

$f (x) = {(e^{x} + e^{- x})}^{2}$

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.

$F (x) = \int {(e^{x} + e^{- x})}^{2} d x$

Step 4

Simplify.

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Step 4.1

Rewrite ${(e^{x} + e^{- x})}^{2}$ as $(e^{x} + e^{- x}) (e^{x} + e^{- x})$ .

$\int (e^{x} + e^{- x}) (e^{x} + e^{- x}) d x$

Step 4.2

Expand $(e^{x} + e^{- x}) (e^{x} + e^{- x})$ using the FOIL Method.

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Step 4.2.1

Apply the distributive property.

$\int e^{x} (e^{x} + e^{- x}) + e^{- x} (e^{x} + e^{- x}) d x$

Step 4.2.2

Apply the distributive property.

$\int e^{x} e^{x} + e^{x} e^{- x} + e^{- x} (e^{x} + e^{- x}) d x$

Step 4.2.3

Apply the distributive property.

$\int e^{x} e^{x} + e^{x} e^{- x} + e^{- x} e^{x} + e^{- x} e^{- x} d x$

Step 4.3

Simplify and combine like terms.

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Step 4.3.1

Simplify each term.

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Step 4.3.1.1

Multiply $e^{x}$ by $e^{x}$ by adding the exponents.

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Step 4.3.1.1.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int e^{x + x} + e^{x} e^{- x} + e^{- x} e^{x} + e^{- x} e^{- x} d x$

Step 4.3.1.1.2

Add $x$ and $x$ .

$\int e^{2 x} + e^{x} e^{- x} + e^{- x} e^{x} + e^{- x} e^{- x} d x$

Step 4.3.1.2

Multiply $e^{x}$ by $e^{- x}$ by adding the exponents.

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Step 4.3.1.2.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int e^{2 x} + e^{x - x} + e^{- x} e^{x} + e^{- x} e^{- x} d x$

Step 4.3.1.2.2

Subtract $x$ from $x$ .

$\int e^{2 x} + e^{0} + e^{- x} e^{x} + e^{- x} e^{- x} d x$

Step 4.3.1.3

Simplify $e^{0}$ .

$\int e^{2 x} + 1 + e^{- x} e^{x} + e^{- x} e^{- x} d x$

Step 4.3.1.4

Multiply $e^{- x}$ by $e^{x}$ by adding the exponents.

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Step 4.3.1.4.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int e^{2 x} + 1 + e^{- x + x} + e^{- x} e^{- x} d x$

Step 4.3.1.4.2

Add $- x$ and $x$ .

$\int e^{2 x} + 1 + e^{0} + e^{- x} e^{- x} d x$

Step 4.3.1.5

Simplify $e^{0}$ .

$\int e^{2 x} + 1 + 1 + e^{- x} e^{- x} d x$

Step 4.3.1.6

Multiply $e^{- x}$ by $e^{- x}$ by adding the exponents.

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Step 4.3.1.6.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int e^{2 x} + 1 + 1 + e^{- x - x} d x$

Step 4.3.1.6.2

Subtract $x$ from $- x$ .

$\int e^{2 x} + 1 + 1 + e^{- 2 x} d x$

Step 4.3.2

Add $1$ and $1$ .

$\int e^{2 x} + 2 + e^{- 2 x} d x$

Step 5

Split the single integral into multiple integrals.

$\int e^{2 x} d x + \int 2 d x + \int e^{- 2 x} d x$

Step 6

Let $u_{1} = 2 x$ . 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 6.1

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

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Step 6.1.1

Differentiate $2 x$ .

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

Step 6.1.2

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

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

Step 6.1.3

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

$2 \cdot 1$

Step 6.1.4

Multiply $2$ by $1$ .

$2$

Step 6.2

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

$\int e^{u_{1}} \frac{1}{2} d u_{1} + \int 2 d x + \int e^{- 2 x} d x$

Step 7

Combine $e^{u_{1}}$ and $\frac{1}{2}$ .

$\int \frac{e^{u_{1}}}{2} d u_{1} + \int 2 d x + \int e^{- 2 x} d x$

Step 8

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

$\frac{1}{2} \int e^{u_{1}} d u_{1} + \int 2 d x + \int e^{- 2 x} d x$

Step 9

The integral of $e^{u_{1}}$ with respect to $u_{1}$ is $e^{u_{1}}$ .

$\frac{1}{2} (e^{u_{1}} + C) + \int 2 d x + \int e^{- 2 x} d x$

Step 10

Apply the constant rule.

$\frac{1}{2} (e^{u_{1}} + C) + 2 x + C + \int e^{- 2 x} d x$

Step 11

Let $u_{2} = - 2 x$ . 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 11.1

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

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Step 11.1.1

Differentiate $- 2 x$ .

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

Step 11.1.2

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

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

Step 11.1.3

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

$- 2 \cdot 1$

Step 11.1.4

Multiply $- 2$ by $1$ .

$- 2$

Step 11.2

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

$\frac{1}{2} (e^{u_{1}} + C) + 2 x + C + \int e^{u_{2}} \frac{1}{- 2} d u_{2}$

Step 12

Simplify.

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Step 12.1

Move the negative in front of the fraction.

$\frac{1}{2} (e^{u_{1}} + C) + 2 x + C + \int e^{u_{2}} (- \frac{1}{2}) d u_{2}$

Step 12.2

Combine $e^{u_{2}}$ and $\frac{1}{2}$ .

$\frac{1}{2} (e^{u_{1}} + C) + 2 x + C + \int - \frac{e^{u_{2}}}{2} d u_{2}$

Step 13

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

$\frac{1}{2} (e^{u_{1}} + C) + 2 x + C - \int \frac{e^{u_{2}}}{2} d u_{2}$

Step 14

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

$\frac{1}{2} (e^{u_{1}} + C) + 2 x + C - (\frac{1}{2} \int e^{u_{2}} d u_{2})$

Step 15

The integral of $e^{u_{2}}$ with respect to $u_{2}$ is $e^{u_{2}}$ .

$\frac{1}{2} (e^{u_{1}} + C) + 2 x + C - \frac{1}{2} (e^{u_{2}} + C)$

Step 16

Simplify.

$\frac{1}{2} e^{u_{1}} + 2 x - \frac{1}{2} e^{u_{2}} + C$

Step 17

Substitute back in for each integration substitution variable.

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Step 17.1

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

$\frac{1}{2} e^{2 x} + 2 x - \frac{1}{2} e^{u_{2}} + C$

Step 17.2

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

$\frac{1}{2} e^{2 x} + 2 x - \frac{1}{2} e^{- 2 x} + C$

Step 18

The answer is the antiderivative of the function $f (x) = {(e^{x} + e^{- x})}^{2}$ .

$F (x) =$ $\frac{1}{2} e^{2 x} + 2 x - \frac{1}{2} e^{- 2 x} + C$