Explanation

(c)
Let $I = \int_{}^{}\frac{x\tan^{- 1}x}{\left( 1 + x^{2} \right)^{\frac{3}{2}}}dx$
Put $x = \tan\theta\ \ \ \Rightarrow dx = \sec^{2}\theta\ d\theta$
$\therefore I = \int_{}^{}\frac{\tan\theta\tan^{- 1}{(\tan\theta)}}{\left( 1 + \tan^{2}\theta \right)^{\frac{3}{2}}}\sec^{2}\theta\ d\theta$
$\Rightarrow I = \int_{}^{}\frac{\theta\tan\theta\sec^{2}\theta}{\sec^{3}\theta}d\theta = \int_{}^{}\theta\sin\theta\ d\theta$
$\Rightarrow I = \theta\left( - \cos\theta \right) + \int_{}^{}{\cos\theta}\ d\theta$
$= - \theta\cos\theta + \sin\theta + c$
$\Rightarrow \ I = - \tan^{- 1}{x.\frac{1}{\sqrt{1 + x^{2}}} +}\frac{x\ }{\sqrt{1 + x^{2}}} + c$
$\Rightarrow \ I = \frac{(x - \tan^{- 1}x)}{\sqrt{1 + x^{2}}} + c$