Biết

D

- Hướng dẫn giải

Đáp án D

$$\begin{gathered} u=\ln \left(x+1\right)\Rightarrow du=\frac{1}{x+1}dx \\ dv=2dx \Rightarrow v=x^2 \\ \Rightarrow I= x^2 \ln \left(x+1\right)\overline{)\begin{array}{c}2\\ 0\end{array}-{\int }_0^2\frac{x^2}{x+1}dx}=4\ln 3-{\int }_0^2(x-1+\frac{1}{x+1})dx \\ =4\ln 3 - \left(\frac{x^2}{2}-x+\ln \left(x+1\right)\right)\overline{)\begin{array}{c}2\\ 0\end{array}=3\ln 3 \to \left\{\begin{array}{l}a=3\\ b=3\end{array}\right.} \\ \Rightarrow 6a+7b=39 \end{gathered}$$