Hi. I'm stuck on this integral problem:

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

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

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- Feb 11th 2013, 08:50 PMinfraRedintegral with e
Hi. I'm stuck on this integral problem:

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

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, 09:15 PMchiroRe: 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, 09:27 PMProve ItRe: integral with e
$\displaystyle \displaystyle \begin{align*} \int{\frac{85}{e^{-x} + 1}\,dx} &= \int{\frac{85e^x}{e^x \left( e^{-x} + 1 \right)} \, dx} \\ &= \int{\frac{85e^x}{1 + e^x}\,dx} \end{align*}$

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

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