Explanation

(a)
Let $I = \int_{}^{}{\sin^{- 1}\left( \frac{2x}{1 + x^{2}} \right)dx}$
Put $x = \tan{\theta\ \Rightarrow \ dx = \sec^{2}{\theta\ d\theta}}$
$\therefore\ \ \ I = \int_{}^{}{\sin^{- 1}{\left( \sin{2\theta} \right).\sec^{2}{\theta\ d\theta}}}$
$= 2\int_{}^{}\theta\sec^{2}\theta d\theta = 2\lbrack\theta\tan\theta - \int_{}^{}{\tan\theta}d\theta\rbrack$
$= 2\left\lbrack \theta\tan\theta + \log{\cos\theta} \right\rbrack + c$
$= 2\left\lbrack x\tan^{- 1}x + \log\frac{1}{\sqrt{1 + x^{2}}} \right\rbrack + c$
$= 2x\tan^{- 1}x - \log\left( 1 + x^{2} \right) + c$
$= f(x) - \log\left( 1 + x^{2} \right) + c$ [given]
$\therefore\ f(x) = 2x\tan^{- 1}x$