Integrating Reciprocated Exponential Function!

Calculus Level 2

$\large \int \dfrac1{1 + e^{-x}} \, dx = \, ?$

Clarification : $C$ denotes the arbitrary constant of integration .

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

\ln(1 - e^x) + C

\ln(1 + e^x) + C

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

2 solutions

Yash Dev Lamba
Mar 16, 2016

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

Take $1+e^x=t$ $\implies$ $e^xdx=dt$

$\int\frac{e^{x}}{1+e^{x}}dx=\int\frac{1}{t}dt=ln|t|+c=ln(1+e^{x})+c$

Nice solution! My method would've taken longer.

Kirubel Solomon - 5 years, 3 months ago

how did you get to that first equality?

Tiago Lima - 5 years, 2 months ago

since $e^{-x}=\frac{1}{e^{x}}$

Yash Dev Lamba - 5 years, 2 months ago

Chew-Seong Cheong
Mar 16, 2016

Similar solution and my presentation is as follows:

$\begin{aligned} \int \frac{1}{1+e^{-x}} dx & = \int \frac{e^x}{e^x + 1} dx \\ & = \int \frac{d}{\color{#D61F06}{dx}} \ln (1+e^x) \color{#D61F06}{dx} \\ & = \int d (\ln (1+e^x)) \\ & = \boxed{\ln (1+e^x) + C} \end{aligned}$

Integrating Reciprocated Exponential Function!

2 solutions

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