If $\int \frac{2 x+3}{(x-1)\left(x^2+1\right)} d x$

$$ =\log _e\left\{(x-1)^{\frac{5}{2}}\left(x^2+1\right)^2\right\}-\frac{1}{2} \tan ^{-1} x+\mathrm{A} $$

where A is an arbitrary constant, then the value of $a$ is

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

$$ \int \frac{x+\sin x}{1+\cos x} d x= $$

If $\int \tan ^4 x \mathrm{~d} x=\mathrm{a} \tan ^3 x+\mathrm{b} \tan x+\mathrm{c} x+\mathrm{k}$ (where k is the constant of integration) then the value of $\mathrm{a}-\mathrm{b}+\mathrm{c}=$