Tích phân $I=\int\limits_{1}^{3}{\frac{3+\ln x}{{{(x+1)}^{2}}}dx}$ bằng A. $3+\ln \frac{27}{16}.$ B. $\frac{1}{4}(3-\ln \frac{27}{16}).$ C. $\frac{1}{4}(3+\ln \frac{27}{16}).$ D. $\frac{1}{4}\ln \frac{27}{16}.$
Đáp án đúng: C Đặt $\left\{ \begin{array}{l}u=3+\ln x\\dv=\frac{1}{{{(x+1)}^{2}}}dx\end{array} \right.=>\left\{ \begin{array}{l}du=\frac{1}{x}dx\\v=-\frac{1}{x+1}\end{array} \right..$