Calculus Examples

$\frac{e^{x}}{e^{x} + 1}$

Step 1

Write $\frac{e^{x}}{e^{x} + 1}$ as a function.

$f (x) = \frac{e^{x}}{e^{x} + 1}$

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.

$F (x) = \int \frac{e^{x}}{e^{x} + 1} d x$

Step 4

Let $u = e^{x} + 1$ . Then $d u = e^{x} d x$ , so $\frac{1}{e^{x}} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.1

Let $u = e^{x} + 1$ . Find $\frac{d u}{d x}$ .

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Step 4.1.1

Differentiate $e^{x} + 1$ .

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

Step 4.1.2

By the Sum Rule, the derivative of $e^{x} + 1$ with respect to $x$ is $\frac{d}{d x} [e^{x}] + \frac{d}{d x} [1]$ .

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

Step 4.1.3

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

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

Step 4.1.4

Differentiate using the Constant Rule.

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Step 4.1.4.1

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$e^{x} + 0$

Step 4.1.4.2

Add $e^{x}$ and $0$ .

$e^{x}$

Step 4.2

Rewrite the problem using $u$ and $d u$ .

$\int \frac{1}{u} d u$

Step 5

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

$\ln (| u |) + C$

Step 6

Replace all occurrences of $u$ with $e^{x} + 1$ .

$\ln (| e^{x} + 1 |) + C$

Step 7

The answer is the antiderivative of the function $f (x) = \frac{e^{x}}{e^{x} + 1}$ .

$F (x) =$ $\ln (| e^{x} + 1 |) + C$