Nếu đặt u = ln(x+2) và dv = xdx thì tích phân

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