∫^2_1(x dx)/((x+1)(x+2))=

The correct answer is option (3) → $\log\frac{32}{27}$

$\int\limits^{2}_{1}\frac{x}{(x+1)(x+2)}dx=\int\limits^{2}_{1}\frac{2(x+1)-(x+2)}{(x+1)(x+2)}dx$

$=\int\limits^{2}_{1}\frac{2}{(x+2)}-\frac{1}{(x+1)}dx=\left|\log\frac{(x+2)^2}{(x+1)}\right|^{2}_{1}$

$=\log\frac{32}{27}$