Question

Eavaluate : $\int x{\tan }^{-1}x$

• A. $\frac{1}{2}(x^2+1){\tan }^{-1}x-\frac{1}{2}x+C$
• B. $\frac{1}{2}{x}^2{\tan }^{-1}x-\frac{1}{2}x+C$
• C. $\frac{1}{2}{x}^2{\tan }^{-1}x+\frac{1}{2}x+C$
• D. none of these

Answer 1

The correct option is D none of these

$I=\int x{\tan }^{-1}xdx$

We know that, $\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

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

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

$=\frac{1}{2}{x}^2{\tan }^{-1}x-\frac{1}{2}\int (1-\frac{1}{x^2+1})dx$

$=\frac{1}{2}{x}^2{\tan }^{-1}x-\frac{1}{2}[x-{\tan }^{-1}x]$

$I=\frac{1}{2}(x^2+1){\tan }^{-1}x-\frac{1}{2}x+C$