what is d integral of sqrt(tanx)

?v(tan x) dx
Let tan x = t^2
=> sec^2 x dx = 2t dt
=> dx = [2t / (1 + t^4)]dt
=> Integral
= ? 2t^2 / (1 + t^4) dt
= ?[(t^2 + 1) + (t^2 - 1)] / (1 + t^4) dt
= ?(t^2 + 1) / (1 + t^4) dt + ?(t^2 - 1) / (1 + t^4) dt
= ?(1 + 1/t^2) / (t^2 + 1/t^2) dt + ?(1 - 1/t^2) / (t^2 + 1/t^2) dt
= ?(1 + 1/t^2)dt / [(t - 1/t)^2 + 2] + ?(1 - 1/t^2)dt / [(t + 1/t)^2 -2]
Let t - 1/t = u for the first integral => (1 + 1/t^2)dt = du
and t + 1/t = v for the 2nd integral => (1 - 1/t^2)dt = dv
Integral
= ?du/(u^2 + 2) + ?dv/(v^2 - 2)
= (1/v2) arctan (u/v2) + (1/2v2) ln l(v -v2)/(v + v2)l + c
= (1/v2) arctan [(t^2 - 1)/tv2] + (1/2v2) ln l (t^2 + 1 - tv2) / t^2 + 1 + tv2) + c
= (1/v2) arctan [(tanx - 1)/(v2tan x)] + (1/2v2) ln l [tanx + 1 - v(2tan x)] / [tan x + 1 + v(2tan x)] + c