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

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

$$\int \cos ^{\frac{-3}{7}} x \cdot \sin ^{\frac{-11}{7}} x d x=$$

$$\int \frac{\mathrm{e}^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] \mathrm{d} x,$$ where $x>0$ is