$\int \frac{1}{(x-1) \sqrt{x^2-1}} d x$ equals
Answer & explanation
Correct answer: option 4
Let
$I=\int \frac{1}{(x-1) \sqrt{x^2-1}} d x=\int \frac{1}{(x-1)^{3 / 2} \sqrt{x+1}} d x$
$\Rightarrow I=\int\left(\frac{x+1}{x-1}\right)^{3 / 2} \frac{1}{(x+1)^2} d x=\frac{1}{2} \int\left(\frac{x-1}{x+1}\right)^{-3 / 2} d\left(\frac{x-1}{x+1}\right)$
$\Rightarrow I=-\left(\frac{x-1}{x+1}\right)^{-1 / 2}+C=-\sqrt{\frac{x+1}{x-1}}+C$