I want to evaluate $ \int \frac{1}{x} \sqrt{\frac{x+a}{x-a}}\mathrm dx $.
$ x=a\cosh(2t), \int \frac{1}{x} \sqrt{\frac{x+a}{x-a}}dx= \int \frac{2\tanh(2t)}{\tanh(t)}dt= \int \frac{4}{1+\tanh^2(t)}\mathrm dt $
$ u=\tanh(t), \int \frac{4}{1+\tanh^2(t)}dt=2 \int (\frac{1}{1+u^2}+\frac{1}{1-u^2})\mathrm du=2 \mathrm{artanh}(u)+2\arctan(u)+C= $ $ \mathrm{arcosh}(x/a)+2\arctan(\sqrt{\frac{x^2-a^2}{x^2+a^2}})+C $
However, I could not manage to show that the derivative of this function is $ \frac{1}{x} \sqrt{\frac{x+a}{x-a}} $.