Question

$\int \frac{1}{\left(x-1\right)\sqrt{x^2+1}}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{dx}{\left(x-1\right)\sqrt{x^2+1}} \\ \mathrm{Putting}x-1=\frac{1}{t} \\ \Rightarrow dx=-\frac{1}{t^2}dt \\ \therefore I=\int \frac{-{\displaystyle \frac{1}{t^2}}dt}{\left({\displaystyle \frac{1}{t}}\right)\sqrt{{\left(1+{\displaystyle \frac{1}{t}}\right)}^2+1}} \\ =\int \frac{-{\displaystyle \frac{1}{t}}dt}{\sqrt{1+{\displaystyle \frac{1}{t^2}}+{\displaystyle \frac{2}{t}}+1}} \\ =\int \frac{-{\displaystyle \frac{1}{t}}dt}{{\displaystyle \frac{\sqrt{t^2+1+2t+t^2}}{t}}} \\ =\int \frac{-dt}{\sqrt{2t^2+2t+1}} \\ =-\frac{1}{\sqrt{2}}\int \frac{dt}{\sqrt{t^2+t+{\displaystyle \frac{1}{2}}}} \\ =-\frac{1}{\sqrt{2}}\int \frac{dt}{\sqrt{t^2+t+{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{2}}}} \\ =-\frac{1}{\sqrt{2}}\int \frac{dt}{\sqrt{{\left(t+{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{1}{2}}\right)}^2}} \\ =-\frac{1}{\sqrt{2}}\log \left|t+\frac{1}{2}+\sqrt{{\left(t+\frac{1}{2}\right)}^2+\frac{1}{4}}\right|+\mathrm{C}\mathrm{where}t=\frac{1}{x-1} \end{gathered}$$