Chọn: Đáp án A

Ta có:$I=\int\limits_{0}^{1}{\frac{3x}{{{\left( x+1 \right)}^{2}}}dx+2\int\limits_{0}^{1}{\frac{\ln \left( 3x+1 \right)}{{{\left( x+1 \right)}^{2}}}}}dx$

Đặt $u=\ln \left( 3x+1 \right)\Rightarrow du=\frac{3dx}{{{\left( x+1 \right)}^{2}}}\Rightarrow v=-\frac{1}{x+1}$

Áp dụng công thức tích phân từng phần ta có

$I=\int\limits_{0}^{1}{\frac{3x}{{{\left( x+1 \right)}^{2}}}dx}-\frac{2\ln {{\left( 3x+1 \right)}^{2}}}{x+1}\left| \begin{align}  & 1 \\  & 0 \\ \end{align} \right.+6\int\limits_{0}^{1}{\left( \frac{dx}{\left( 3x+1 \right)\left( x+1 \right)} \right)}$

$=\int\limits_{0}^{1}{\left( \frac{3}{x+1}-\frac{3}{{{\left( x+1 \right)}^{2}}} \right)dx}-\ln 4+\int\limits_{0}^{1}{\left( \frac{9}{3x+1}-\frac{3}{x+1} \right)dx}$

$=\left( 3\ln \left| x+1 \right|+\frac{3}{x+1} \right)\left| \begin{align}  & 1 \\  & 0 \\ \end{align} \right.-2\ln 2+\int\limits_{0}^{1}{\left( \frac{9}{3x+1}-\frac{3}{x+1} \right)dx=-\frac{3}{2}+\ln 2+\int\limits_{0}^{1}{\left( \frac{9}{3x+1}-\frac{3}{x+1} \right)dx}}$$\Rightarrow \left\{ \begin{align}  & a=9 \\  & b=3 \\ \end{align} \right.$ Nháp:

$6\int\limits_{0}^{1}{\frac{dx}{\left( 3x+1 \right)\left( x+1 \right)}=6\int\limits_{0}^{1}{\left( \frac{m}{3x+1}+\frac{n}{x+1} \right)dx}}$ . tìm $m,n$ . Ta có:$m\left( x+1 \right)+n\left( 3x+1 \right)=1$

$\left\{ \begin{align}  & x=-1\Rightarrow n=-\frac{1}{2} \\  & x=0\Rightarrow m+n=1\Rightarrow m=\frac{3}{2} \\ \end{align} \right.$$\Rightarrow 6\int\limits_{0}^{1}{\frac{dx}{\left( 3x+1 \right)\left( x+1 \right)}=6\int\limits_{0}^{1}{\left( \frac{3}{2\left( 3x+1 \right)}-\frac{1}{2\left( x+1 \right)} \right)dx=\int\limits_{0}^{1}{\left( \frac{9}{3x+1}-\frac{3}{x+1} \right)dx}}}$