How is the integral $\int \frac{1}{ae^{mx}+be^{-mx}} \, dx$ solved? I am trying to determine how the following integral was solved, or at least the name of article deriving this solution:
$$
\int \frac{1}{ae^{mx}+be^{-mx}} \, dx
= \frac{1}{m\sqrt{ab}}\,\tan^{-1} \left(e^{mx}\sqrt{\frac{a}{b}}\,\right)$$

I don't know of the article where this result comes from, but it is definitely doable without it! First, notice that equality will fail if $a$ or $b$ are equal to zero. You cannot calculate the integral at all if both $a$ and $b$ equal zero. But the integral is very easy to solve if either $a$ or $b$ is zero. If both are non-zero, then you can pattern match what you have to something that we can integrate. Remember that $$\int \frac{dy}{1+y^2} = \arctan(y)+C$$ for any function $y$. Let's try to rewrite your integral to look like the above integral.  $$\int{\frac{1}{ae^{mx}+be^{-mx}}}dx =  \int \frac{1}{be^{-mx}}\left(\frac{1}{\frac{a}{b}e^{2mx}+1}\right)dx \\  =\frac{1}{b}\int \frac{e^{mx}}{\left(\sqrt{\frac{a}{b}e^{2mx}}\right)^2+1}dx \\  =\frac{1}{b}\int \frac{e^{mx}}{\left(\sqrt{\frac{a}{b}}e^{mx}\right)^2+1}dx$$ Introduce $$y = \sqrt{\frac{a}{b}}e^{mx}$$ to obtain $$dy = m\sqrt{\frac{a}{b}}e^{mx}dx \implies dx = \frac{1}{m}\sqrt{\frac{b}{a}}e^{-mx}dy$$ Hence $$\frac{1}{b}\int \frac{e^{mx}}{\left(\sqrt{\frac{a}{b}}e^{mx}\right)^2+1}dx = \frac{1}{b}\int \frac{e^{mx}}{\left(y\right)^2+1} \left(\frac{1}{m}\sqrt{\frac{b}{a}}e^{-mx}dy \right) \\ = \frac{1}{m}\sqrt{\frac{b}{a}}\cdot \frac{1}{b}\int \frac{e^{mx}\cdot e^{-mx}}{\left(y\right)^2+1}dy \\ = \frac{1}{m\sqrt{ab}}\int \frac{1}{\left(y\right)^2+1}dy$$ The integral has successfully been rewritten into a form that we recognize and can integrate. Thus, $$ \frac{1}{m\sqrt{ab}}\int \frac{1}{\left(y\right)^2+1}dy= \frac{1}{m\sqrt{ab}}\left(\arctan(y)+C \right)\\ = \frac{1}{m\sqrt{ab}}\arctan(y)+\tilde{C} \\ = \frac{1}{m\sqrt{ab}}\arctan\left(\sqrt{\frac{a}{b}}e^{mx} \right)+\tilde{C} $$
