Question

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

• A. $\frac{x^2}{2}{\tan }^{-1}x+\frac{1}{2}x+\frac{1}{2}{\tan }^{-1}x+c$
• B. $\frac{x^2+1}{2}{\tan }^{-1}x-\frac{x}{2}+c$
• C. $(x^2+1){\tan }^{-1}x+\frac{x}{2}+c$
• D. $(x^2+1){\tan }^{-1}x-\frac{x}{2}+c$

Answer 1

Answer: (B)

$\frac{x^2+1}{2}{\tan }^{-1}x-\frac{x}{2}+c$

$\int x{\tan }^{-1}xdx$
$x=\tan \theta$
$\int \theta \tan \theta se{c}^2\theta d\theta$
$\int \theta \tan \theta se{c}^2\theta d\theta =\frac{\theta {\tan }^2\theta }{2}+\int \left(\frac{{\tan }^2\theta }{2}\right)d\theta$
$=\frac{\theta {\tan }^2\theta }{2}-\frac{1}{2}\int (se{c}^2\theta -1)d\theta$
$=\frac{\theta {\tan }^2\theta }{2}-\frac{1}{2}\tan \theta +\frac{1}{2}\theta +c$
$\frac{x^2}{2}{\tan }^{-1}x-\frac{1}{2}x+\frac{1}{2}{\tan }^{-1}x+c$
$\frac{x^2+1}{2}{\tan }^{-1}x-\frac{x}{2}+c$