I=∫√tan x dx

Let tan x = t

So, sec^2 x dx = 2 t dt

So, (1+tan^2 x) dx = 2 t dt

So, dx = (2 t dt) / (1+ t^4)

Substituting in I, eventually I becomes

∫ {(t^2+1 ) + (t^2-!)} / (t^4 +1)

On further simplifying,

I= ∫ (t^2+!) / (t^4+1) dt + ∫ (t^2-1) / (t^4 +1) dt

I=∫(1+ 1/t^2) / (t^2 + !/t^2) dt + ∫ (1- 1/t^2) / (t^2+ 1/t^2) dt

For da 1st integral sub t- 1/t =u

So,da 1st integral becums ∫du / (u^2 + 2)

Further simplifyn, it becums 1/√2 tan‾1{(t – 1/t) / √2}

For 2nd integral, sub t+ 1/t =v

So,da 2nd interal becums ∫dv / (v^2 -2)

Now solv both da integrals ………...........

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Ans is

(1/√2) tan‾1 {(√tan x - √cot x +√2 ) / √2} + (-1/ 2√2) log {(√tan x +√cot x +√2) / (√tan x + √cot x - √2)} + c