Biết

D

- Hướng dẫn giải

Đáp án D

Đặt $\left\{\begin{array}{l}u=\ln \left(x+1\right)\\ dv=2xdx\end{array}\right.\Rightarrow \left\{\begin{array}{l}du=\frac{1}{x+1}\\ v=x^2\end{array}\right.\Rightarrow {\int }_0^{2}2x\ln \left(x+1\right)dx={\overline{)\left[x^2\ln \left(x+1\right)\right]}}_0^2-{\int }_0^2\frac{x^2}{x+1}dx$ 

${\overline{)\left[x^2\ln \left(x+1\right)\right]}}_0^2-{\int }_0^2\left(x-1+\frac{1}{x+1}\right)dx={\overline{)\left[x^2\ln \left(x+1\right)\right]}}_0^2-{\overline{)\left[\frac{x^2}{2}-x+\ln \left(x+1\right)\right]}}_0^2=3\ln 3\Rightarrow \left\{\begin{array}{l}a=3\\ b=3\end{array}\right.$$\Rightarrow 6a+7b=39$.