Using the identity, sinhx = 1/2(e^x-e^-x) and the substitution v=e^-x to find the integral of cosechx.
Hello,
$\displaystyle \text{cosech}(x)=\frac{1}{\sinh(x)}$
that should finish the preliminaries
you'll have $\displaystyle 2 \int \frac{1}{e^x-e^{-x}} ~dx$
substitute, as you're told, $\displaystyle v=e^{-x}$ (note that $\displaystyle e^x=\frac 1v$) and show your work...
You're making this too hard. "Fractions" is not a good way to think of this. For instance integrating $\displaystyle \frac{1}{x^2+2x+5}$ is integrating a fraction but is very complex. In your problem you have variables over a constant. Notice that $\displaystyle \frac{abc....z}{2}=\frac{1}{2} abc...z$. And you should know that any constant in front of an integral can be brought outside of the integral and applied to the answer after integration. So for this one, your constant is 1/2, bring it out and apply it after.
So like Moo said, v = e^{-x}, thus 1/v = e^{x}. Now use v-substitution. So we have this expression, $\displaystyle \frac{2}{\frac{1}{v}-v}$ If v=e^{-x}, $\displaystyle \ln(v)=-x$. Thus $\displaystyle \frac{1}{v}dv=-dx$. The new integral is now $\displaystyle -\int \frac{2}{\frac{1}{v}-v}\frac{1}{v}dv=\frac{2}{1-v^2}dv$. Are you with me so far?