If $$\int_\limits2^e\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] d x=a+\frac{b}{\log 2}$$, then

$$\int_\limits0^{\pi / 4} \log (1+\tan x) d x=$$

If $$\int_\limits0^{\frac{\pi}{2}} \frac{d x}{5+4 \sin x}=A \tan ^{-1} B$$, then A + B =

$$\int_0^a \sqrt{\frac{x}{a-x}} d x=$$