Question

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

Answer 1

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