u substitution solving indefinite integrals

[mlane]

I have an integral of (e^x-e^-x)/(e^x+e^-x) *dx
I am not sure how to do the substitution.
If choose u=e^x+e^-x I know the derivative of this is the top part of the equation but I am lost on the meaning of what I am trying to do. How do I know when I am done.

I get lost with the implicit part of differentiation.
would it be du/dx=e^x-e^-x.
I don't know what to do with it from here because I don't understand what I need to get to in the end.

 

[Gene]

Yes d(e^x+e^-x) = (e^x-e^-x)dx so you have du/u which you should recognize as d(ln(u)). The integral is
ln(u)+ln(C) = ln(C*u). Just un-sub and you are done.

 

[galactus]

You could possibly start by noticing that \(\displaystyle \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}=tanh(x)=\frac{sinh(x)}{cosh(x)}\)

Let \(\displaystyle u=cosh(x)\) and \(\displaystyle du=sinh(x)\)

\(\displaystyle \int\frac{du}{u}=ln(u)\)

Since u=cosh(x), we have ln(cosh(x))+C

\(\displaystyle cosh(x)=\frac{e^{x}+e^{-x}}{2}\)

\(\displaystyle ln(\frac{e^{x}+e^{-x}}{2})+C\)