If $$\int \frac{\sin x}{3+4 \cos ^2 x} \mathrm{~d} x=\mathrm{A} \tan ^{-1}(\mathrm{~B} \cos x)+\mathrm{c}$$, (where $$\mathrm{c}$$ is a constant of integration), then the value of $$\mathrm{A}+\mathrm{B}$$ is

$$\int(\sqrt{\tan x}+\sqrt{\cot x}) d x=$$

Let $$\alpha \in\left(0, \frac{\pi}{2}\right)$$ be fixed. If the integral $$\int \frac{\tan x+\tan \alpha}{\tan x-\tan \alpha} \mathrm{d} x=\mathrm{A}(x) \cos 2 \alpha+\mathrm{B}(x) \sin 2 \alpha+\mathrm{c},$$ (where $$\mathrm{c}$$ is a constant of integration), then functions $$\mathrm{A}(x)$$ and $$\mathrm{B}(x)$$ are respectively

$$\int \frac{x+1}{x\left(1+x \mathrm{e}^x\right)^2} \mathrm{~d} x=$$