Mathematics - Indefinite Integration Question with Solution | TestHub

Let $\mathrm{I}=\int x{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx$ $$\therefore \mathrm{I}=2{\int }_{\text{II }}x\cdot {\tan }^{-1}xdx$$ Applying Integration by parts $$\begin{array}{r}\mathrm{I}=2\left[{\tan }^{-1}x\int xdx-\int \left(\frac{d}{dx}\left({\tan }^{-1}x\right)\int xdx\right)dx\right]\\ \mathrm{I}=2\left[\frac{x^2}{2}{\tan }^{-1}x-\int \frac{1}{1+x^2}\times \frac{x^2}{2}dx\right]+c\\ \mathrm{I}=2\left[\frac{x^2}{2}{\tan }^{-1}x-\frac{1}{2}\int \frac{x^2+1-1}{x^2+1}dx\right]+c\\ \mathrm{I}=2\left[\frac{x^2}{2}{\tan }^{-1}x-\frac{1}{2}\int \frac{x^2+1}{x^2+1}dx+\frac{1}{2}\int \frac{1}{1+x^2}dx\right]+c\end{array}$$ $$\begin{array}{rl}\mathrm{I}& =2\left[\frac{x^2}{2}{\tan }^{-1}x-\frac{1}{2}\int 1.dx+\frac{1}{2}{\tan }^{-1}x\right]+c\\ \mathrm{I}& =2\left[\frac{x^2}{2}{\tan }^{-1}x-\frac{x}{2}+\frac{1}{2}{\tan }^{-1}x\right]+c\\ \mathrm{I}& =x^2{\tan }^{-1}x+{\tan }^{-1}x-x+c\\ & \text{ or }I=-x+\left(x^2+1\right){\tan }^{-1}x+c\end{array}$$