Evaluation of $\int_{-2}^{-1} \tan^{-1}\sqrt{\frac{x+1}{x-1}} dx$ Evaluation of $$\int_{-2}^{-1} \tan^{-1}\sqrt{\frac{x+1}{x-1}} dx$$
requires a special care when done by hand. The question is: How to do it and what is the right answer.

Note that when $x\in (-\infty, -1] \cup (1,\infty)$,
$$\frac{d}{dx} \tan^{-1} \sqrt{\frac{x+1}{x-1}}=-\frac{1}{2|x|} \frac{1}{\sqrt{x^2-1}}.$$
Next the integration by parts leads to  $$I=\int_{-2}^{-1} \tan^{-1} \sqrt{\frac{x+1}{x-1}}.~1 ~dx=\left . x \tan^{-1} \sqrt{\frac{x+1}{x-1}}\right|_{-2}^{-1}-\frac{1}{2}\int_{-2}^{-1} \frac{dx}{\sqrt{x^2-1}}.$$
$$\implies I=\frac{\pi}{3}-\frac{1}{2} \int_{1}^{2} \frac{dz}{\sqrt{z^2-1}}=\frac{\pi}{3}-\ln\sqrt{2+\sqrt{3}}.$$
