Question

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

Answer 1

$$\begin{gathered} \text{We have,} \\ I=\int x\sqrt{\frac{1-x}{1+x}}dx \\ \Rightarrow I=\int x\sqrt{\frac{\left(1-x\right)\left(1-x\right)}{\left(1+x\right)\left(1-x\right)}}dx \\ \Rightarrow I=\int x\frac{\left(1-x\right)}{\sqrt{1-x^2}}dx \\ \Rightarrow I=\int \frac{x-x^2}{\sqrt{1-x^2}}dx \\ \Rightarrow I=\int \frac{x-x^2-1+1}{\sqrt{1-x^2}}dx \\ \Rightarrow I=\int \frac{-x^2+1}{\sqrt{1-x^2}}dx+\int \frac{x-1}{\sqrt{1-x^2}}dx \\ \Rightarrow I=\int \sqrt{1-x^2}dx+\int \frac{x}{\sqrt{1-x^2}}dx-\int \frac{1}{\sqrt{1-x^2}}dx \\ \Rightarrow I=\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}\left(x\right)+C_1-\sqrt{1-x^2}+C_2-{\sin }^{-1}\left(x\right)+C_3\left[\because \int \frac{x}{\sqrt{1-x^2}}dx=-\sqrt{1-x^2}+C_2\right] \\ \Rightarrow I=\sqrt{1-x^2}\left(\frac{x}{2}-1\right)-\frac{1}{2}{\sin }^{-1}\left(x\right)+C \end{gathered}$$