Explanation

(d)
Let $I = \int_{}^{}{\tan^{- 1}{\begin{matrix}
x \\
I
\end{matrix}.\begin{matrix}
1 \\
II
\end{matrix}}}\ dx$
$= \tan^{- 1}x.x - \int_{}^{}\frac{1}{1 + x^{2}}.x\ dx$
$= x\tan^{- 1}x - I_{1}$ …(i)
Where $I_{1} = \int_{}^{}\frac{x}{1 + x^{2}}dx$
Put $1 + x^{2} = t \Rightarrow x\ dx = \frac{1}{2}dt$
$\therefore\ I_{1} = \frac{1}{2}\int_{}^{}\frac{1}{t}\ dt = \frac{1}{2}\log t$
$= \frac{1}{2}\log{|1 + x^{2}|}$
$\therefore$ From eq . (i), we get
$I = x\tan^{- 1}x - \frac{1}{2}\log{|1 + x^{2}|} + c$