Calculus Examples

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

Step 1

Write $\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$ as a function.

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

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.

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

Step 4

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 4.1

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

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Step 4.1.1

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

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

Step 4.1.2

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 4.1.3

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 4.1.4

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

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Step 4.1.4.1

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 4.1.4.1.1

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 4.1.4.1.2

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 4.1.4.1.3

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

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

Step 4.1.4.2

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 4.1.4.3

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 4.1.4.4

Multiply $- 1$ by $1$ .

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

Step 4.1.4.5

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

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

Step 4.1.4.6

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

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

Step 4.2

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

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

Step 5

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

$\ln (| u_{2} |) + C$

Step 6

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

$\ln (| e^{x} + e^{- x} |) + C$

Step 7

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

$F (x) =$ $\ln (| e^{x} + e^{- x} |) + C$