# Integral Help Please

I am having a heck of a hard time with this integral... I have tried everything what I can think of:

$\int \! \left( {e^{x}}+{e^{-x}} \right) ^{-1}{dx}$

I tried integration by parts... I ended up getting $\left( {e^{x}} \right) ^{-1}$ even thought the right answer, according to Maple and my graphing calculator is $\arctan \left( {e^{x}} \right)$

I then tried using substitution... I made $u={e^{x}}$ and then ${\it du}={e^{x}}{\it dx}$ but that doesn't help me, because I don't have ${e^{x}}{\it dx}$ but rather I have ${\frac {{\it dx}}{{e^{x}}}}.$

Can anyone point me in the right direction? Thanks!

## Answers and Replies

make this substituion u=e^x
then (e^x)dx=du ---> udx=du ---> dx=du/u
the int becomes (1/(u+1/u))*(du/u) ---> (1/(1+u^2))*du the int is then arctan = arctan[e^x]
sorry but i don't knwo this latex language to write it in a more elegant way.

Thank you so much! It worked :) .