Evaluate: $\int \sqrt {\tan (x)} \,dx$
My Attempt: $$=\int \sqrt {\tan (x)} \,dx$$ Let $u=\tan (x)$ $$du=\sec^2 (x) \,dx$$ Then $$=\int \frac {\sqrt {u}}{u^2+1} \,du$$ Let, $s=\sqrt u$ and $ds=\dfrac 1 {2\sqrt u} \, du$ So, $$=2\int \dfrac {s^2}{s^4+1} \, ds$$
What to do further?
The rest it's the following. $$\frac{s^2}{s^4+1}=\frac{s^2}{s^4+2s^2+1-2s^2}=\frac{1}{2\sqrt2}\left(\frac{s}{s^2-\sqrt2s+1}-\frac{s}{s^2+\sqrt2s+1}\right)$$ and you get $\ln$ and $\arctan$.