If $I=∫(\sqrt{tanx}+\sqrt{cotx})dx $ = f(x) + c then f(x) is equal to

$\sqrt{2}tan^{-1}(\frac{tanx-1}{\sqrt{2}\sqrt{tanx}})$

Let $I=∫(\sqrt{tanx}+\sqrt{cotx})dx=∫\sqrt{2}\frac{sinx+cosx}{\sqrt{2sinxcosx}}dx$ 

If  sin x – cos x =  p then (cos x  + sin x) ⇒  dx = dp so that 

$I=\sqrt{2}∫\frac{dp}{\sqrt{1-p^2}}=\sqrt{2}sin^{-1}p=\sqrt{2}sin^{-1}(sinx-cosx)$

$=\frac{\pi}{\sqrt{2}}-\sqrt{2}cos^{-1}(sinx-cosx)$

$=\sqrt{2}tan^{-1}\frac{sinx-cosx}{\sqrt{1-(sinx-cosx)^2}}=\sqrt{2}tan^{-1}\frac{sinx-cosx}{\sqrt{2sinxcosx}}$

$=\sqrt{2}tan^{-1}(\frac{tanx-1}{\sqrt{2}\sqrt{tanx}})$

Hence (A), (B), (C) are correct answers.