How do you find int(e^-x)/(1+e^-x)dx?

4 Answers
Jun 28, 2018

inte^(-x)/(e^(-x)+1)dx=x-ln(e^x+1)+c

Explanation:

inte^(-x)/(e^(-x)+1)dx=int(-(e^(-x)+1)')/(e^(-x)+1)dx=-int((e^(-x)+1)')/(e^(-x)+1)dx=-ln(e^(-x)+1)+c

  • ln(e^(-x)+1)=ln(1/e^x+1)=ln((e^x+1)/e^x)=ln(e^x+1)-lne^x=ln(e^x+1)-x

Therefore, inte^(-x)/(e^(-x)+1)dx=-(ln(e^x+1)-x)+c=x-ln(e^x+1)+c

, cinRR

Jun 28, 2018

The answer is =-ln(1+e^-x)+C

Explanation:

Perform this integral by substitution

Let u=1+e^-x

=>, du=-e^-xdx

Therefore, the integral is

I=int(e^-xdx)/(1+e^-x)

=-int(du)/u

=-ln(u)

=-ln(1+e^-x)+C

Jun 28, 2018

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

Explanation:

Let u=e^{-x}. This yields du = -e^{-x}dx

Rewrite

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

Substitute -e^{-x}dx=du and e^{-x}=u:

- \int \frac{1}{1+u}du

This integral is -ln(1+u)+C.

Substitute back u=e^{-x} to get

\int \frac{e^{-x}}{1+e^{-x}}dx = -ln(1+e^{-x})+C

Jun 28, 2018

I tried this:

Explanation:

Have a look:
enter image source here