Showing equivalence of the hyperbolic function I've been thinking about this problem for about 30 minutes now and still I don't have a clue on what I'd do next. I have this equation and I aim to show equivalence (proof) between
$$\int\operatorname{csch}{(u)}\,\mathrm{d}u=\ln{\left(\tanh{\frac{u}{2}}\right)}+c.$$
If I made a move on the right hand side and placed $\tanh{\frac{u}{2}}$ with $\frac{\cosh{x}-1}{\sinh{x}}$, what can I do next? Any help or guide on answering this question is very much appreciated. :)
 A: Hint: The main identity to know about hyperbolic functions is the hyperbolic-Pythagorean theorem (just as how the classical Pythagorean theorem is the most important relation for trig functions):
$$1=\cosh^2{(\phi)}-\sinh^2{(\phi)}.$$
Now, using the definition of the function $\operatorname{csch}{(z)}$ and a little algebra, we have:
$$\begin{align}
\int\operatorname{csch}{(u)}\,\mathrm{d}u
&=\int\frac{1}{\sinh{(u)}}\,\mathrm{d}u\\
&=\int\frac{1}{\sinh{(u)}}\cdot\frac{\sinh{(u)}}{\sinh{(u)}}\,\mathrm{d}u\\
&=\int\frac{\sinh{(u)}}{\sinh^2{(u)}}\,\mathrm{d}u\\
&=\int\frac{\sinh{(u)}}{\cosh^2{(u)}-1}\,\mathrm{d}u\\
&=\int\frac{1}{x^2-1}\,\mathrm{d}x,
\end{align}$$
where in the last line we've substituted $x=\cosh{(u)}$, $dx=\sinh{(u)}\,du$.

Hint #2: The last integral can be broken down into elementary integrals via partial fractions:
$$\frac{1}{x^2-1}=\frac{1}{2(x-1)}-\frac{1}{2(x+1)},\\
\begin{align}~~~~~~~~~~~~~\implies\int\frac{\mathrm{d}x}{x^2-1}&=\frac12\int\frac{\mathrm{d}x}{x-1}-\frac12\int\frac{\mathrm{d}x}{x+1}\\
&=\frac12\ln{(x-1)}-\frac12\ln{(x+1)}+\color{grey}{\text{constant}}\\
&=\frac12\ln{\left(\frac{x-1}{x+1}\right)}+\color{grey}{\text{constant}}.
\end{align}$$
Hence, after back-substituting $x=\cosh{(u)}$, 
$$\int\operatorname{csch}{(u)}\,\mathrm{d}u=\frac12\ln{\left(\frac{\cosh{(u)}-1}{\cosh{(u)}+1}\right)}+\color{grey}{\text{constant}}.$$
This solves the integral. The last step of the problem asks us to show that $\ln{\left(\tanh{\left(\frac{u}{2}\right)}\right)}$ is also an anti-derivative of $\operatorname{csch}{(u)}$, which is just a matter of algebra.

Edit
The partial fraction decomposition is found through a purely algebraic process (and has nothing to do with differentiation). Start by writing the rational function as a linear combination of its linear factors:
$$\frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}.$$
Multiplying by $(x-1)(x+1)$, we have:
$$1=A(x+1)+B(x-1).$$
To find the values of $A$ and $B$, first set $x=1$, and the equation reduces to $1=2A$. Similarly, by setting $x=-1$, we find $1=-2B$.
A: Hint: Use the fact that $\text{csch }x=\dfrac1{\sinh x}$ and $\sinh x=\dfrac{e^x-e^{-x}}2$ . Then let $t=e^x$. The desired result soon follows.
