Take the tour ×  Find $$\int \sqrt{\tan x}dx$$  My attempt:
$$\text{Let}\ I=\int \sqrt{\tan(x)}dx$$  $$\text{Let}\
u=\tan(x), du=(1+\tan^{2}(x))dx$$  $$I=\int
\frac{\sqrt{u}}{u^{2}+1}$$  $$\text{Let}\ v=\sqrt{u},
dv=\frac{du}{2\sqrt{u}}$$  $$I=2\int \frac{v^{2}}{v^{4}+1}$$
$$\text{Let}\ t=\frac{1}{v} \therefore dt=\frac{-dv}{v^2}$$
$$\therefore I=\int
\frac{\frac{1}{t^2}}{1+\frac{1}{t^4}}\times\frac{-dt}{t^2}$$
$$I=-\int \frac{dt}{1+t^4}$$  Where do I go from here?
share|improve this question Could you please tell me how I can
know whether there is a duplicate before I post? I will go
through the thread and see if the answer I am looking for is
there. Thank you. –  User79217 May 9 at 9:57 I changed the
question. I do not think it is a duplicate now. –  User79217 May
9 at 10:04 You are asking how to compute the integral in the
last line. See accepted answer to that question. It is
defenitely a duplicate –  Norbert May 9 at 10:12 It is always
difficult to determine whether your question have been asked be