Nếu đặt

Câu hỏi: Nếu đặt $\left\{\begin{array}{l}u=\ln (x+2)\\ dv=xdx\end{array}\right.$ thì tích phân $I={\int }_0^{1}x.\ln (x+2)dx$ trở thành:

A. $I=\frac{x^2\ln \left(x+2\right)}{2}{|}_0^1-\frac{1}{2}{\int }_0^1\frac{x^2}{x+2}dx$

B. $I=x^2\ln \left(x+2\right){|}_0^1-\frac{1}{4}{\int }_0^1\frac{x^2}{x+2}dx$

C. $I=\frac{x^2\ln \left(x+2\right)}{2}{|}_0^1+{\int }_0^1\frac{x^2}{x+2}dx$

D. $I=\frac{x^2\ln \left(x+2\right)}{4}{|}_0^1-\frac{1}{4}{\int }_0^1\frac{x^2}{x+2}dx$