Tích phân $I=\int\limits_{0}^{2}{\dfrac{x+4}{{{x}^{2}}+3x+2}}\text{dx}=a\ln 3+b\ln 2$. Khi đó ${{b}^{2}}-a$ bằng bao nhiêu?

Ta có $I=\int\limits_{0}^{2}{\dfrac{x+4}{{{x}^{2}}+3x+2}}\text{dx}=\int\limits_{0}^{2}{\left( \dfrac{3}{x+1}-\dfrac{2}{x+2} \right)\text{dx}}=3\int\limits_{0}^{2}{\dfrac{1}{x+1}\text{dx}}-2\int\limits_{0}^{2}{\dfrac{1}{x+2}\text{dx}}$
$=3\left. \ln \left| x+1 \right| \right|_{0}^{2}\left. -2\ln \left| x+2 \right| \right|_{0}^{2}$ $=3\ln 3-2\ln 4+2\ln 2=3\ln 3-2\ln 2$
Vậy $a=3,b=-2$ nên ${{b}^{2}}-a=1$.