Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{1+x^2}{\sqrt{1-x^2}}dx \\ =\int \frac{2-\left(1-x^2\right)}{\sqrt{1-x^2}}dx \\ =2\int \frac{1}{\sqrt{1-x^2}}dx-\int \frac{1-x^2}{\sqrt{1-x^2}}dx \\ =2\int \frac{1}{\sqrt{1-x^2}}dx-\int \sqrt{1-x^2}dx \\ =2{\sin }^{-1}x-\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x\right]+C \\ =2{\sin }^{-1}x-\frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}{\sin }^{-1}x+C \\ =\frac{3}{2}{\sin }^{-1}x-\frac{x}{2}\sqrt{1-x^2}+C \end{gathered}$$