# integral with e

• Feb 11th 2013, 09:50 PM
infraRed
integral with e
Hi. I'm stuck on this integral problem:

$\int{\frac{85}{e^{-x}+1}\,dx}$

I keep playing around with substitution and +/- tricks and such like, and I just don't seem to be figuring it out.
• Feb 11th 2013, 10:15 PM
chiro
Re: integral with e
Hey infraRed.

Hint: Try multiplying the integral inside by x/x and use the substitution u = e^(-x) + 1.
• Feb 11th 2013, 10:27 PM
Prove It
Re: integral with e
Quote:

Originally Posted by infraRed
Hi. I'm stuck on this integral problem:

$\int{\frac{85}{e^{-x}+1}\,dx}$

I keep playing around with substitution and +/- tricks and such like, and I just don't seem to be figuring it out.

\displaystyle \begin{align*} \int{\frac{85}{e^{-x} + 1}\,dx} &= \int{\frac{85e^x}{e^x \left( e^{-x} + 1 \right)} \, dx} \\ &= \int{\frac{85e^x}{1 + e^x}\,dx} \end{align*}

Now make the substitution \displaystyle \begin{align*} u = e^x \implies du = e^x\,dx \end{align*} and the integral becomes

\displaystyle \begin{align*} \int{\frac{85e^x}{1 + e^x}\,dx} &= \int{\frac{85}{1 + u}\,du} \\ &= 85\ln{\left| 1 + u \right|} + C \\ &= 85\ln{\left| 1 + e^x \right|} + C \\ &= 85\ln{ \left( 1 + e^x \right) } + C \textrm{ since } 1 + e^x > 0 \textrm{ for all } x \end{align*}