How Do You Solve This Tricky Integral?

[Weather Freak]

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!

 

[ziad1985]

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.

 

[Weather Freak]

Thank you so much! It worked :) .