Encontrar a $$\int \sqrt{\tan x}dx$ $
Mi intento:
$$\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}$$
¿A dónde voy de aquí?