Explanation

(d)
Let $I = \int_{}^{}\frac{1}{\left( \sin x + 4 \right)(\sin x - 1)}dx$
$= \frac{1}{5}\int_{}^{}\frac{(\sin{x + 4)} - (\sin x - 1)}{\left( \sin x + 4 \right)(\sin x - 1)}dx$
$= \frac{1}{5}\int_{}^{}\frac{1}{(\sin{x - 1)}}dx - \frac{1}{5}\int_{}^{}\frac{1}{\left( \sin x + 4 \right)}dx$
$= \frac{1}{5}\int_{}^{}\frac{\sec^{2}\frac{x}{2}}{2\tan\frac{x}{2} - 1 - \tan^{2}\frac{x}{2}}dx - \frac{1}{5}\int_{}^{}\frac{\sec^{2}\frac{x}{2}}{2\tan\frac{x}{2} + 4 + 4\tan^{2}\frac{x}{2}}dx$
Put $\tan{\frac{x}{2} = t}$
$\Rightarrow \sec^{2}\frac{x}{2}dx = 2t\ dt$
$\therefore I = \frac{1}{5}\int_{}^{}\frac{2dt}{2t - 1 - t^{2}} - \frac{1}{5}\int_{}^{}\frac{2dt}{\lbrack 2t + 4\left( 1 + t^{2} \right)\rbrack}$
$\therefore\ I = - \frac{2}{5}\int_{}^{}\frac{dt}{2t - 1 - t^{2}} - \frac{1}{10}\int_{}^{}\frac{dt}{t^{2} - \frac{1}{2}t + 1}$
$= \frac{- 2}{5}\int_{}^{}\frac{1}{(t - 1)^{2}}dt - \frac{1}{10}\int_{}^{}\frac{1}{\left( t + \frac{1}{4} \right)^{2}\left( \frac{\sqrt{15}}{4} \right)^{2}}$
$= \frac{2}{5}\ \frac{1}{(t - 1)} - \frac{2}{5\sqrt{15}}\tan^{- 1}\left( \frac{4t + 1}{\sqrt{15}} \right) + c$
$= \frac{2}{5}\ \frac{1}{(\tan\frac{x}{2} - 1)} - \frac{2}{5\sqrt{15}}\tan^{- 1}\left( \frac{4\tan\frac{x}{2} + 1}{\sqrt{15}} \right) + c$
$\therefore A = \frac{2}{5},\ B = \frac{- 2}{5\sqrt{15}},\ f(x) = \frac{4\tan\frac{x}{2} + 1}{\sqrt{15}}$