I would like to solve the following primitive:
$$\int\frac{1-a\cosh(x)}{(\cosh(x)-a)^2}\,dx$$
where $a$ is a constant, $0\leq a\leq1$.
I really don't know how to start. I can't relate the numerator with the denominator. There's no $\sinh(x)$ in the integrand, and I can't see any similarity with the well known hyperbolic functions derivatives. How would you do it?
Here is a slightly more structured approach to arriving at an answer which is not simply based, as you say, on a "pure guess."
Let $$J = \int \frac{1 - a \cosh x}{(\cosh x - a)^2} \, dx.$$
To find the integral $J$, the appearance of the all squared term in the denominator of the integrand suggests it may be possible to make use of the so-called reverse quotient rule (For another example using this method see here).
Recall that if $u$ and $v$ are differentiable functions, from the quotient rule $$\left (\frac{u}{v} \right )' = \frac{u' v - v' u}{v^2},$$ and it is immediate that $$\int \frac{u' v - v' u}{v^2} \, dx = \int \left (\frac{u}{v} \right )' \, dx = \frac{u}{v} + C. \tag1$$
For the integral $J$ we see that $v = \cosh x - a$. So $v' = \sinh x$. Now for the (generally) harder bit but in this particular case it is quite easy. We need to find a function $u(x)$ such that $$u' v - v' u = u'(\cosh x - a) - u \sinh x = 1 - a \cosh x.$$ After a little trial and error it is quite easy to see that if $$u = \sinh x,$$ as $$u' = \cosh x,$$ this gives $$u' v - v' u = \cosh^2 x - a \cosh x - \sinh^2 x = 1 - a \cosh x,$$ where we have used $\cosh^2 x - \sinh^2 x = 1$, as required.
Our integral can now be readily found. The result is: \begin{align} J &= \int \left (\frac{\sinh x}{\cosh x - a} \right )' \, dx = \frac{\sinh x}{\cosh x - a} + C. \end{align} as expected.
Let's start by conjecturing an antiderivative of the form $\frac{f(x)}{\cosh x-a}$, so $$f^\prime(x)(\cosh x-a)-f(x)\sinh x=1-a\cosh x=\cosh^2x-a\cosh x-\sinh^2x.$$By inspection, one solution is $f(x)=\sinh x.$
Hint: Differentiate $$\frac{\sinh(x)}{\cosh(x)-a}+C$$ with respect to $x$
