Đáp án và lời giải
Đáp án:C
Lời giải:\begin{eqnarray*} \displaystyle\int\limits_1^3 \dfrac{3+\ln x}{(x+1)^2}\mathrm{\,d}x&=&\displaystyle\int\limits_1^3 \dfrac{3}{(x+1)^2}\mathrm{\,d}x +\displaystyle\int\limits_1^3 \dfrac{\ln x}{(x+1)^2}\mathrm{\,d}x= \left.-\dfrac{3}{x+1}\right|_1^3-\displaystyle\int\limits_1^3 \ln x\mathrm{\,d}\left(\dfrac{1}{x+1}\right)\\ &=&-3\left(\dfrac{1}{4}-\dfrac{1}{2}\right)-\left(\dfrac{1}{x+1}\cdot \ln x-\displaystyle\int\limits_1^3 \dfrac{1}{x+1}\mathrm{\,d} \ln x \right)\\ &=&\dfrac{3}{4}-\dfrac{1}{4}\ln3 + \displaystyle\int\limits_1^3 \dfrac{1}{(x+1)x}\mathrm{\,d}x\\ &=&\dfrac{3}{4}-\dfrac{1}{4} \ln3 + \displaystyle\int\limits_1^3 \left(\dfrac{1}{x}-\dfrac{1}{x+1}\right)\mathrm{\,d}x=\dfrac{3}{4}-\dfrac{1}{4}\ln3 +\ln x \big|_1^3-\ln(x+1) \big|_1^3\\ &=&\dfrac{3}{4}-\dfrac{1}{4}\ln3+\ln3-(\ln4-\ln2)=\dfrac{3}{4}(1+\ln3)-\ln2. \end{eqnarray*} Suy ra $a=\dfrac{3}{4}$, $b=1\Rightarrow a^2+b^2=\dfrac{9}{16}+1=\dfrac{25}{16}$.
