The value of the integral $\int_1^2 \frac{x \mathrm{~d} x}{(x+2)(x+3)}$ is

$$ \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{x}{1+\sin x} d x= $$

$$ \int_0^2 \frac{3 x+1}{x^2+4} d x= $$

$$ \int_{-1}^3\left(\tan ^{-1}\left(\frac{x}{x^2+1}\right)+\tan ^{-1}\left(\frac{x^2+1}{x}\right)\right) d x= $$