Question

$\int {\tan }^{-1}\left(\sqrt{x}\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int {\tan }^{-1}\sqrt{x}dx \\ =\int \frac{\sqrt{x}.{\tan }^{-1}\sqrt{x}dx}{\sqrt{x}} \\ \mathrm{Let}\sqrt{x}=t \\ \Rightarrow \frac{1}{2\sqrt{x}}dx=dt \\ =\frac{dx}{\sqrt{x}}=2dt \\ \therefore I=2\int \underset{\mathrm{II}}{t}.\underset{\mathrm{I}}{{\tan }^{-1}\left(t\right)}dt \\ =2\left[{\tan }^{-1}t\int tdt-\int \left\{\frac{d}{dt}\left({\tan }^{-1}t\right)\int tdt\right\}dt\right] \\ =2\left[{\tan }^{-1}\left(t\right).\frac{t^2}{2}-\int \frac{1}{1+t^2}.\frac{t^2}{2}dt\right] \\ ={\tan }^{-1}\left(t\right).t^2-\int \frac{t^2}{1+t^2}dt \\ ={\tan }^{-1}\left(t\right).t^2-\int \left(\frac{1+t^2-1}{1+t^2}\right)dt \\ ={\tan }^{-1}\left(t\right).t^2-\int dt+\int \frac{dt}{1+t^2} \\ ={\tan }^{-1}\left(t\right).t^2-t+{\tan }^{-1}\left(t\right)+C\left(\because \sqrt{x}=t\right) \\ ={\tan }^{-1}\left(\sqrt{x}\right).x-\sqrt{x}+{\tan }^{-1}\sqrt{x}+C \\ =\left(x+1\right){\tan }^{-1}\sqrt{x}-\sqrt{x}+C \end{gathered}$$