$\int (\sqrt{\tan x}+\sqrt{\cot x})dx=$

$I=\int (\sqrt{\tan x}+\sqrt{\cot x})dx=$

$=\int \frac{\sqrt{\sin x}}{\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\sin x}}dx$

$=\int \frac{\sin x+\cos x}{\sqrt{sinx.\cos x}}dx=\sqrt{2}\int \frac{\sin x+\cos x}{\sqrt{2\sin x\cos x}}dx$

Put $\sin x - \cos x=t\Rightarrow \left(\cos x+\sin x\right)dx=dt$

$and t^2={\left(\sin x-\cos x\right)}^2=1-2\sin x\cos x$

$I=\sqrt{2}\int \frac{dt}{\sqrt{1-t^2}}=\sqrt{2}{\sin }^{-1}t+c$

$\Rightarrow \sqrt{2}{\sin }^{-1}(\sin x-\cos x)+c$