∫√tan x dx = ?

Evaluate ∫√tan x dx .

Let u = √tan x
x = tan^(– 1) u²
dx = (2u/(u^4 + 1)) du

∴∫√tan x dx = ∫(2u²/(u^4 + 1)) du

Since u^4 + 1 = u^4 + 2u² + 1 – 2u²
　　　　　　= (u² + 1)² – (√2 u)²
　　　　　　= (u² – √2 u + 1)(u² + √2 u + 1)
So we let 2u²/(u^4 + 1) ≡ (Au + B)/(u² – √2 u + 1) + (Cu + D)/(u² + √2 u + 1)
2u² ≡ (Au + B)(u² + √2 u + 1) + (Cu + D)(u² – √2 u + 1)
2u² ≡ Au³ + √2 Au² + Au + Bu² + √2 Bu + B + Cu³ – √2 Cu² + Cu + Du² – √2 Du + D
2u² ≡ (A + C)u³ + (√2 A + B – √2 C + D)u² + (A + √2 B + C – √2 D)u + B + D
　　╭
　　│A + C = 0……………………....(1)
∴─┤√2 A + B – √2 C + D = 2………(2)
　　│A + √2 B + C – √2 D = 0………(3)
　　│B + D = 0………………………(4)
　　╰
Put (1) into (3) and (4) into (2) ,
√2 A – √2 C = 2
A – C = √2…………(5)
√2 B – √2 D = 0
B – D = 0…………...(6)

Solving (1) , (4) , (5) and (6) , we get A = (√2)/2 , B = 0 , C = – (√2)/2 and D = 0

∴∫(2u²/(u^4 + 1)) du
= ((√2)/2) ∫(u/(u² – √2 u + 1)) du – ((√2)/2) ∫(u/(u² + √2 u + 1)) du
= ((√2)/4) ∫(2u/(u² – √2 u + 1)) du – ((√2)/4) ∫(2u/(u² + √2 u + 1)) du
= ((√2)/4) ∫((2u – √2)/(u² – √2 u + 1)) du + (1/2) ∫(du/(u² – √2 u + 1)) – ((√2)/4) ∫((2u + √2)/(u² + √2 u + 1)) du + (1/2) ∫(du/(u² + √2 u + 1))
= ((√2)/4) ∫((d(u² – √2 u + 1))/(u² – √2 u + 1)) – ((√2)/4) ∫((d(u² + √2 u + 1))/(u² + √2 u + 1)) + (1/2) ∫(du/((u + (√2)/2)² + 1/2)) + (1/2) ∫(du/((u – (√2)/2)² + 1/2))

Let u + (√2)/2 = ((√2)/2) tan θ , u – (√2)/2 = ((√2)/2) tan φ
du = ((√2)/2) sec² θ dθ , du = ((√2)/2) sec² φ dφ

∴((√2)/4) ∫((d(u² – √2 u + 1))/(u² – √2 u + 1)) – ((√2)/4) ∫((d(u² + √2 u + 1))/(u² + √2 u + 1)) + (1/2) ∫(du/((u + (√2)/2)² + 1/2)) + (1/2) ∫(du/((u – (√2)/2)² + 1/2))
= ((√2)/4) ln (u² – √2 u + 1) – ((√2)/4) ln (u² + √2 u + 1) + (1/2) ∫([((√2)/2) sec² θ dθ]/[(((√2)/2) tan θ)² + 1/2]) + (1/2) ∫([((√2)/2) sec² φ dφ]/[(((√2)/2) tan φ)² + 1/2]) + C_1
= ((√2)/4) ln ((u² – √2 u + 1)/(u² + √2 u + 1)) + (1/2) ∫([((√2)/2) sec² θ dθ]/[(1/2) sec² θ]) + (1/2) ∫([((√2)/2) sec² φ dφ]/[(1/2) sec² φ]) + C_1
= ((√2)/4) ln ((u² – √2 u + 1)/(u² + √2 u + 1)) + (1/2) ∫√2 dθ + (1/2) ∫√2 dφ + C_1
= ((√2)/4) ln ((u² – √2 u + 1)/(u² + √2 u + 1)) + (√2 θ)/2 + (√2 φ)/2 + C
= ((√2)/4) ln ((u² – √2 u + 1)/(u² + √2 u + 1)) + (tan^(– 1) (√2 u + 1) + tan^(– 1) (√2 u – 1))/√2 + C
= ((√2)/4) ln ((tan x – √(2 tan x) + 1)/(tan x + √(2 tan x) + 1)) + (tan^(– 1) (√(2 tan x) + 1) + tan^(– 1) (√(2 tan x) – 1))/√2 + C

You really want to find it? The steps are very cubersome.
I show you the answer here first. See if you're really interested.

∫√tan x dx

= 2 ^ -3/2 { -2 tan^-1[ 1 - (2^1/2) tan^(1/2) (x) ] + 2 tan^-1[ ( 2^(1/2) tan^(-1/2) (x) + 1 ] + ln[ - tan(x) + 2^(1/2) tan^(1/2) (x) - 1 ] - ln[ tan(x) + 2^(1/2) tan^(1/2) (x) + 1 ] }


Differentiate this expression with respect to x, we can get √tan x

Pardon me if I have typo.