# Thread: Integrate inverse of cosh

1. ## Integrate inverse of cosh

I have an equation i wish to integrate

$
\int\frac{dx}{coshx+1}
$

What is the best way to start a problem like this. Should i substitute or integrate by parts? Any help appreciated!!

2. Try using its e identity.

$cosh(x)=\frac{1}{2}\left(e^{x}+e^{-x}\right)$

$\frac{1}{\frac{1}{2}\left(e^{x}+e^{-x}\right)+1}=$

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

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

3. but $\frac{2e^{x}}{e^{2x}+2e^{x}+1}=\frac{2e^{x}}{\left ( e^{x}+1 \right)^{2}}.$

4. I don't understand how you got to

Originally Posted by galactus

$\frac{2e^{x}}{e^{2x}+2e^{x}+1}=\frac{2}{e^{x}+1}-\frac{2}{(e^{x}+1)^{2}}$
from this

Originally Posted by galactus

$\frac{1}{\frac{1}{2}\left(e^{x}+e^{-x}\right)+1}$

5. It's just a little algebra.

$\frac{2u}{(u+1)^{2}}$

$\frac{2u+2-2}{(u+1)^{2}}$

$\frac{2(u+1)-2}{(u+1)^{2}}$

$\frac{2}{u+1}-\frac{2}{(u+1)^{2}}$

Now, plug the e back in.