Ta có
$\frac{x^2+\ln(x^2e^x)}{(x+2)^2}=\frac{x^2+\ln(x^2)+\ln e^x}{(x+2)^2}=\frac{x^2+2\ln x+x}{(x+2)^2}$
Suy ra
$I = \int\limits_{1}^{2}\frac{x^2+\ln(x^2e^x)}{(x+2)^2}dx= I_1+ I_2$. Trong đó
$\bullet\quad I_1=\int\limits_{1}^{2}\frac{x^2+x}{(x+2)^2}dx$
$I_1 = \int\limits_{1}^{2}\frac{(x+2)^2+2-3(x+2)}{(x+2)^2}dx= \int\limits_{1}^{2}\left ( 1+\frac{2}{(x+2)^2}-\frac{3}{x+2} \right )dx=\left[ {x-\frac{2}{x+2}-3\ln(x+2) } \right]_{1}^{2}=\frac{7}{6}-\ln\frac{64}{27}$
$\bullet\quad I_2=\int\limits_{1}^{2}\frac{2\ln x}{(x+2)^2}dx$. Ta dùng phương pháp tích phân từng phần.
$I_2 = -2\int\limits_{1}^{2}\ln xd\left ( \frac{1}{x+2} \right )=-2\left[ { \frac{\ln x}{x+2} } \right]_{1}^{2} +\int\limits_{1}^{2}\frac{2}{x(x+2)}dx=\left[ {\frac1x-\frac{1}{x+2}- \frac{2\ln x}{x+2}} \right]_{1}^{2}=\frac{1}{2}\ln\frac{9}{8}$.