Nasty inverse tan integral I've been trying the following integral, which Mathematica will happily do:
$$\int\frac{1}{\sqrt{x^2(1-cx)(x-1)}}\mathrm{d}x = \tan ^{-1}\left(\frac{c x+x-2}{2 \sqrt{x-1} \sqrt{1-c x}}\right) $$
However, some hints as to how to do it by hand would be very helpful. Particularly how you would do it without working back from the answer. I've tried $u=\sqrt{x-1}$ and $x=t^2$ followed by $t=\sec\theta$ but neither worked well. It's been a while since I've done one of these!

Taking $x$ common from $(1-cx),(x-1)$ we get $\int\frac{dx}{\sqrt{x^2.x(\frac{1}{x}-c)x(1-\frac{1}{x})}}$ let $\frac{1}{x}=u$ then $\frac{-dx}{x^2}=du$ our integral changes to $\int\frac{du}{\sqrt{(u-c)(1-u)}}$ . With some manipulation you can find this integral.
