Question

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

Answer 1

$$\begin{gathered} \int \frac{dx}{\sqrt{{\tan }^{-1}x}\left(1+x^2\right)} \\ \mathrm{Let}{\tan }^{-1}x=t \\ \Rightarrow \frac{1}{1+x^2}=\frac{dt}{dx} \\ \Rightarrow \frac{1}{1+x^2}dx=dt \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\int \frac{dx}{\sqrt{{\tan }^{-1}x}\left(1+x^2\right)} \\ =\int \frac{dt}{\sqrt{t}} \\ =\int t^{-\frac{1}{2}}dt \\ =\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}+\mathrm{C} \\ =2\sqrt{t}+\mathrm{C} \\ =2\sqrt{{\tan }^{-1}x}+\mathrm{C} \end{gathered}$$