Calculus Examples

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

Step 1

Factor $2$ out of $2 e^{x} - 2 e^{- x}$ .

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Step 1.1

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

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

Step 1.2

Factor $2$ out of $- 2 e^{- x}$ .

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

Step 1.3

Factor $2$ out of $2 (e^{x}) + 2 (- e^{- x})$ .

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

Step 2

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

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

Step 3

Let $u_{2} = e^{x} + e^{- x}$ . Then $d u_{2} = (e^{x} - e^{- x}) d x$ , so $\frac{1}{e^{x} - e^{- x}} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 3.1

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

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Step 3.1.1

Differentiate $e^{x} + e^{- x}$ .

$\frac{d}{d x} [e^{x} + e^{- x}]$

Step 3.1.2

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

$\frac{d}{d x} [e^{x}] + \frac{d}{d x} [e^{- x}]$

Step 3.1.3

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

$e^{x} + \frac{d}{d x} [e^{- x}]$

Step 3.1.4

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

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Step 3.1.4.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x$ .

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Step 3.1.4.1.1

To apply the Chain Rule, set $u_{1}$ as $- x$ .

$e^{x} + \frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [- x]$

Step 3.1.4.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$e^{x} + e^{u_{1}} \frac{d}{d x} [- x]$

Step 3.1.4.1.3

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

$e^{x} + e^{- x} \frac{d}{d x} [- x]$

Step 3.1.4.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$e^{x} + e^{- x} (- \frac{d}{d x} [x])$

Step 3.1.4.3

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

$e^{x} + e^{- x} (- 1 \cdot 1)$

Step 3.1.4.4

Multiply $- 1$ by $1$ .

$e^{x} + e^{- x} \cdot - 1$

Step 3.1.4.5

Move $- 1$ to the left of $e^{- x}$ .

$e^{x} - 1 \cdot e^{- x}$

Step 3.1.4.6

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$e^{x} - e^{- x}$

Step 3.2

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

$2 \int \frac{1}{{u_{2}}^{2}} d u_{2}$

Step 4

Apply basic rules of exponents.

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Step 4.1

Move ${u_{2}}^{2}$ out of the denominator by raising it to the $- 1$ power.

$2 \int {({u_{2}}^{2})}^{- 1} d u_{2}$

Step 4.2

Multiply the exponents in ${({u_{2}}^{2})}^{- 1}$ .

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Step 4.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$2 \int {u_{2}}^{2 \cdot - 1} d u_{2}$

Step 4.2.2

Multiply $2$ by $- 1$ .

$2 \int {u_{2}}^{- 2} d u_{2}$

Step 5

By the Power Rule, the integral of ${u_{2}}^{- 2}$ with respect to $u_{2}$ is $- {u_{2}}^{- 1}$ .

$2 (- {u_{2}}^{- 1} + C)$

Step 6

Simplify.

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Step 6.1

Rewrite $2 (- {u_{2}}^{- 1} + C)$ as $2 (- \frac{1}{u_{2}}) + C$ .

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

Step 6.2

Simplify.

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Step 6.2.1

Multiply $- 1$ by $2$ .

$- 2 \frac{1}{u_{2}} + C$

Step 6.2.2

Combine $- 2$ and $\frac{1}{u_{2}}$ .

$\frac{- 2}{u_{2}} + C$

Step 6.2.3

Move the negative in front of the fraction.

$- \frac{2}{u_{2}} + C$

Step 7

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

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