Đáp án:
$D$
Giải thích các bước giải:
Ta có: $∫4x(1+lnx)dx$
$=∫(4x+4xlnx)dx$
$=∫4xdx+∫4xlnxdx$
$=4∫xdx+4∫xlnxdx$
$=4.\dfrac{1}{2}x^2+4∫xlnxdx$
$=2x^2+4∫xlnxdx$
Xét $∫xlnxdx$, đặt $\left\{ \begin{array}{l}u=lnx\\dv=xdx\end{array} \right.$
$⇒ \left\{ \begin{array}{l}du=\dfrac{dx}{x}\\v=\dfrac{x^2}{2}\end{array} \right.$
$⇒ ∫xlnxdx=\dfrac{x^2.lnx}{2}-∫\dfrac{x}{2}dx$
$=\dfrac{x^2lnx}{2}-\dfrac{1}{2}.\dfrac{1}{2}x^2$
$=\dfrac{x^2lnx}{2}-\dfrac{x^2}{4}$
Vậy $∫4x(1+lnx)dx=2x^2+4.\Bigg(\dfrac{x^2lnx}{2}-\dfrac{x^2}{4}\Bigg)+C$
$=2x^2+2x^2lnx-x^2+C$
$=2x^2lnx+x^2+C$
