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

If $\mathrm{f}(x)=1+x ; \mathrm{g}(x)=\log x$, then $\int \mathrm{g}(\mathrm{f}(x)) \mathrm{d} x$ is equal to

$$\int \cos (\log x) \mathrm{d} x=$$

$$ \int \frac{2 x+5}{\sqrt{7-6 x-x^2}} d x=A \sqrt{7-6 x-x^2}+B \sin ^{-1}\left(\frac{x+3}{4}\right)+\mathrm{c} $$ (where c is a constant of integration) then the value of $A+B$ is