Question

$\int \frac{x-1}{\sqrt{x^2+1}}dx$

Answer 1

​$$\begin{gathered} \mathrm{Let}I=\int \left(\frac{x-1}{\sqrt{x^2+1}}\right)dx \\ =\int \frac{xdx}{\sqrt{x^2+1}}-\int \frac{dx}{\sqrt{x^2+1}} \\ \mathrm{Putting}{x}^2+1=t \\ \Rightarrow 2xdx=dt \\ \Rightarrow xdx=\frac{dt}{2} \\ \mathrm{Then}, \\ I=\frac{1}{2}\int \frac{dt}{\sqrt{t}}-\int \frac{dx}{\sqrt{x^2+1^2}} \\ =\frac{1}{2}\int t^{-\frac{1}{2}}dt-\int \frac{dx}{\sqrt{x^2+1^2}} \\ =\frac{1}{2}\left[\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}\right]-\int \frac{dx}{\sqrt{x^2+1^2}} \\ =\sqrt{t}-\log \left|x+\sqrt{x^2+1}\right|+C \\ =\sqrt{x^2+1}-\log \left|x+\sqrt{x^2+1}\right|+C \end{gathered}$$