# 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.
• Great, thanks a lot! I followed it up with $t=\sqrt{1-u}$ and then $t=\sqrt{1-c}\sin\theta$ and it worked – dsfkgjn Jun 13 at 15:00