# Integrate

• May 5th 2009, 01:58 PM
tiga killa
Integrate
the intergal of 1/(e^(2x)-1) dx
• May 5th 2009, 02:10 PM
Infophile
Hello,

$\displaystyle u=e^{2x}$ and $\displaystyle du=2u.dx$.

So $\displaystyle \int \frac{1}{1+e^{2x}}dx=\frac{1}{2}\int \frac{1}{u(u-1)}$

But $\displaystyle \frac{1}{u(u-1)}=\frac{1}{u-1}-\frac{1}{u}$.

Hence $\displaystyle \frac{1}{2}\int \left(\frac{1}{u-1}-\frac{1}{u}\right)=\frac{1}{2}\left(\ln(u-1)-\ln(u)\right)+C$

Therefore a primitive is $\displaystyle \color{blue}\boxed{x\mapsto \frac{1}{2}\ln\left(\frac{e^{2x}-1}{e^{2x}}\right)}$ or still $\displaystyle \boxed{x\mapsto \frac{1}{2}\ln(2sh(x))}$

:)
• May 5th 2009, 02:18 PM
skeeter
Quote:

Originally Posted by tiga killa
the intergal of 1/(e^(2x)-1) dx

$\displaystyle \frac{1}{e^{2x}-1} = \frac{1 - e^{2x} + e^{2x}}{e^{2x} - 1} = \frac{e^{2x}}{e^{2x}-1} - \frac{e^{2x}-1}{e^{2x}-1} = \frac{e^{2x}}{e^{2x}-1} - 1$

integrate the last expression
• May 5th 2009, 02:22 PM
Infophile
Yes, why not, it's faster.

Hello skeeter :)