Explanation

(d)
Let $I = \int_{}^{}{\frac{\sqrt{x}}{x + 1}dx} = \int_{}^{}\frac{\sqrt{x}\ dx}{\left( \sqrt{x} \right)^{2} + 1}$
Put $\sqrt{x} = t\ \Rightarrow \ \frac{1}{2\sqrt{x}}dx = dt\ \Rightarrow \ \ \ dx = 2\sqrt{x}dt\$
$\therefore\ I = \int_{}^{}\frac{2t^{2}dt}{t^{2} + 1} = 2\left\lbrack \int_{}^{}\frac{t^{2} + 1}{t^{2} + 1}dt - \int_{}^{}\frac{1}{t^{2} + 1}dt \right\rbrack$
$= 2t - 2\tan^{- 1}{\left( t^{2} + 1 \right) + c}$
$= 2\sqrt{x} - 2\tan^{- 1}{(x + 1)} + c$
But $I = A\sqrt{x} + B\tan^{- 1}{(x + 1)} + c$
$\Rightarrow \ \ A = 2,\ B = - 2$