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

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 $\mathrm{a}-\mathrm{b}+\mathrm{c}=$

$$ \int \frac{x \mathrm{~d} x}{(x-1)(x-2)}= $$

$\int \frac{x^2-4}{x^4+9 x^2+16} \mathrm{dx}=\tan ^{-1}(\mathrm{f}(x))+\mathrm{c}$ (where c is a constant of integration), then value of $f(2)$ is