$$\int \cot x \cdot \log [\log (\sin x)] d x=$$

$$\int \log x \cdot[\log (e x)]^{-2} d x=\ldots$$

If $\int \frac{1}{1-\cot x} d x=A x+B \log |\sin x-\cos x|+c$ then $A+B=\ldots \ldots$

$$\int \frac{d x}{(\sin x+\cos x)(2 \cos x+\sin x)}=$$