If ∫ 1 + tan x sin 2 x d x = A logtan x + B tan x + C then 4 A - B =

$I=\int \frac{1+\sqrt{\tan x}}{\sin 2x}dx=\int \left(\frac{1}{\sin 2x}+\frac{\sqrt{\tan x}}{\sin 2x}\right)dx$

$=\int \frac{1}{2\sin x\cos x}dx+\int \frac{\sqrt{\sin x}}{2\sin x\cos x\sqrt{\cos x}}dx$

$=\int \frac{1}{2{\displaystyle \frac{\sin x}{\cos x}}{\cos }^{2}x}dx+\int \frac{1}{2{\displaystyle \frac{\sqrt{\sin x}}{\sqrt{\cos x}}}{\cos }^{2}x}dx$

$=\int \frac{{\sec }^{2}x}{2{\displaystyle \tan x}}dx+\int \frac{{\sec }^{2}x}{2\sqrt{\tan x}}dx$

$=\frac{1}{2}\ln \left|\tan x\right|+\sqrt{\tan x}+C$

So, $4A-B=4\times \frac{1}{2}-1=1$