Find the following integral: Find $$\int \sqrt{\tan x}dx$$
My attempt:
$$\text{Let}\ I=\int \sqrt{\tan(x)}dx$$
$$\text{Let}\ u=\tan(x), du=(1+\tan^{2}(x))dx$$
$$I=\int \frac{\sqrt{u}}{u^{2}+1}$$
$$\text{Let}\ v=\sqrt{u}, dv=\frac{du}{2\sqrt{u}}$$
$$I=2\int \frac{v^{2}}{v^{4}+1}$$
$$\int_0^\infty\frac{x^2}{1+x^4}dx$$
$$\text{Let}\ t=\frac{1}{v} \therefore dt=\frac{-dv}{v^2}$$
$$\therefore I=\int \frac{\frac{1}{t^2}}{1+\frac{1}{t^4}}\times\frac{-dt}{t^2}$$
$$I=-\int \frac{dt}{1+t^4}$$
Where do I go from here?

First let us compute the following $$\int(\sqrt{\tan x}+\sqrt{\cot x}) dx=\int\frac{\sin x +\cos x}{\sqrt{\sin x\cdot \cos x}}dx=\sqrt 2\int\frac{d(\sin x - \cos x)}{\sqrt{1-(\sin x -\cos x)^2}}$$ which is same as the  $$\sqrt 2\int\frac{dz}{\sqrt{1-z^2}}=\sqrt 2\sin ^{-1}z+c$$ Again compute $$\int(\sqrt{\tan x}-\sqrt{\cot x})dx=\int\frac{\sin x -\cos x}{\sqrt{\sin x\cdot \cos x}}dx=\sqrt 2\int\frac{-d(\sin x+\cos x)}{\sqrt{(\sin x+\cos x)^2 -1} } dx$$ and this is same as $$-\sqrt 2\int \frac {dw}{\sqrt{w^2 -1}}=-\sqrt 2\int\frac{\sec u\tan u}{\tan u} du=-\sqrt 2\ln(\sec u +\tan u)+C$$ where $w=\sec u$. Now add both the integrals to obtain the result.
