∫_1^2(x dx)/(≤ft(x+1)≤ft(x+2)) is equal to

\(\int\limits_{1}^{2}\frac{x}{\left(x+1\right)\left(x+2\right)}dx=\int\limits_{1}^{2}\frac{2(x+1)-(x+2)}{(x+1)(x+2)}dx=\int\limits_{1}^{2}\frac{2}{(x+2)}-\frac{1}{(x+1)}dx\)

$=\left[2\log(x+2)-\log(x+1)\right]_{1}^{2}$

$=2\log 4-\log 3-2\log 3+\log 2$

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