$\int \frac{\sqrt{\tan x}}{\sin x \cos x} dx$ equals :

$2 \sqrt{\tan x}+c$

$I=\int \frac{\sqrt{\tan x}}{\sin x \cos x} \frac{\sqrt{\tan}}{\sqrt{\tan x}} d x$    (Multiplying and dividing by $\sqrt{\tan x}$)

$=\int \frac{\tan x}{\sin x \cos x} \frac{1}{\sqrt{\tan x}} d x$

$=\int \frac{\sin x}{\cos x \sin x \cos x} \frac{1}{\sqrt{\tan x}} d x$

$=\int \frac{1}{\cos ^2 x} \frac{1}{\sqrt{\tan x}} d x$

$=\int \frac{\sec ^2 x d x}{\sqrt{\tan x}}$

let $y=\tan x$

$d y=\sec ^2 x d x$

So  $\int \frac{d y}{\sqrt{y}}=\frac{y^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}= 2 \sqrt{y}=2 \sqrt{\tan x}+c$