# How do you integrate  (1/(e^x+1))dx ?

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29
Nov 19, 2016

Multiply top and bottom by ${e}^{-} x$.

#### Explanation:

$\int \frac{1}{{e}^{x} + 1} \mathrm{dx}$

Divide numerator and denominator by ${e}^{x}$:

$= \int {e}^{-} \frac{x}{1 + {e}^{-} x} \mathrm{dx}$

Rewrite so that the numerator is the derivative of the denominator:

$= - \int \frac{- {e}^{-} x}{1 + {e}^{-} x} \mathrm{dx}$

Integrate directly:

$= - \ln | 1 + {e}^{-} x | + C$

We can leave the answer as is, or simplify further:

$= \ln | \frac{1}{1 + {e}^{-} x} | + C$

$= \ln | {e}^{x} / \left({e}^{x} + 1\right) | + C$

$= x - \ln | {e}^{x} + 1 | + C$

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23
Apr 11, 2017

$\int \frac{\mathrm{dx}}{{e}^{x} + 1} = \int \frac{{e}^{x} + 1 - {e}^{x}}{{e}^{x} + 1} \mathrm{dx} = \int \left(1 - \frac{{e}^{x}}{{e}^{x} + 1}\right) \mathrm{dx} = x - \ln \left({e}^{x} + 1\right)$

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11
Jan 29, 2017

Use the substitution $u = {e}^{x}$ and partial fraction decomposition.

#### Explanation:

$I = \int \frac{1}{{e}^{x} + 1} \mathrm{dx}$

Apply the substitution $u = {e}^{x}$:

$I = \int \frac{1}{u + 1} \left(\frac{\mathrm{du}}{u}\right)$

$\textcolor{w h i t e}{I} = \int \frac{1}{u \left(u + 1\right)} \mathrm{du}$

Apply partial fraction decomposition:

$I = \int \left(\frac{1}{u} - \frac{1}{u + 1}\right) \mathrm{du}$

Integrate term by term:

$I = \ln | u | - \ln | u + 1 | + C$

Reverse the earlier substitution:

$I = x - \ln | {e}^{x} + 1 | + C$

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mason m Share
Mar 25, 2017

Let ${e}^{\frac{x}{2}} = \tan \theta$. Then $\frac{1}{2} {e}^{\frac{x}{2}} \mathrm{dx} = {\sec}^{2} \theta d \theta$.

$\int \frac{\mathrm{dx}}{{e}^{x} + 1} = 2 \int \frac{\frac{1}{2} {e}^{\frac{x}{2}} \mathrm{dx}}{{e}^{\frac{x}{2}} \left({e}^{x} + 1\right)} = 2 \int \frac{{\sec}^{2} \theta d \theta}{\tan \theta \left({\sec}^{2} \theta\right)} = 2 \int \cos \frac{\theta}{\sin} \theta d \theta$

$= 2 \ln \left\mid \sin \right\mid \theta$

From $\tan \theta = {e}^{\frac{x}{2}}$ draw a right triangle to see that $\sin \theta = {e}^{\frac{x}{2}} / \sqrt{{e}^{x} + 1}$:

$= 2 \ln \left\mid {e}^{\frac{x}{2}} / \sqrt{{e}^{x} + 1} \right\mid = \ln \left\mid {e}^{x} / \left({e}^{x} + 1\right) \right\mid = x - \ln \left\mid {e}^{x} + 1 \right\mid + C$

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