If $\int\left(\frac{4 e^x-25}{2 e^x-5}\right) d x=A x+B \log \left(2 e^x-5\right)+c \quad$ (where c is a constant of integration) then

$$\int \tan ^{-1}\left(\frac{1-\sin x}{1+\sin x}\right) d x=$$

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

$\int\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} d x$ equal to