$\int \frac{d x}{(2 x+3) \sqrt{4 x+5}}$ is equal to :

$\tan ^{-1} \sqrt{4 x+5}+c$

Let $I=\int \frac{d x}{(2 x+3) \sqrt{4 x+5}}$

Put $4 x+5=t \Rightarrow x=\frac{t-5}{4}$

$dx=\frac{d t}{4}$

$\Rightarrow I=\frac{1}{4} \int \frac{d t}{\left(\frac{2 t-10}{4}+3\right) \sqrt{t}}=\frac{1}{2} \int \frac{d t}{(t+1) \sqrt{t}}$

Let $\sqrt{t}=u$

∴  $I=\int \frac{d u}{u^2+1}=\tan ^{-1} \sqrt{t}+c$

$\Rightarrow I=\tan ^{-1} \sqrt{4 x+5}+c$

Hence (2) is the correct answer.