$1, I=\int\limits_{1}^{e}=\frac{x^{2}(x.lnx+2)+lnx+1}{xlnx+2}dx=\int\limits_{1}^{e}\left ( x^{2}+\frac{lnx+1}{xlnx+2} \right )dx=\int\limits_{1}^{e}x^{2}dx+\int\limits_{1}^{2}\frac{d(xlnx+2)}{xlnx+2}(d(xlnx+2)=d(xlnx+c)=(lnx+1)dx;c=constVậyI=\frac{x^{3}}{3}+ln(xlnx+2)$ bạn tự thay cận vào nha
$1, I=\int\limits_{1}^{e}=\frac{x^{2}(x.lnx+2)+lnx+1}{xlnx+2}dx=\int\limits_{1}^{e}\left ( x^{2}+\frac{lnx+1}{xlnx+2} \right )dx=\int\limits_{1}^{e}x^{2}dx+\int\limits_{1}^{2}\frac{d(xlnx+1)}{xlnx+2}(d(xlnx+2)=(lnx+1)dx=(lnx+c);c=constVậyI=
$1, I=\int\limits_{1}^{e}=\frac{x^{2}(x.lnx+2)+lnx+1}{xlnx+2}dx=\int\limits_{1}^{e}\left ( x^{2}+\frac{lnx+1}{xlnx+2} \right )dx=\int\limits_{1}^{e}x^{2}dx+\int\limits_{1}^{2}\frac{d(xlnx+
2)}{xlnx+2}
(d(xlnx+2)=
d(
xlnx+
c)=(lnx+
1)
dx;c=const
VậyI=
\frac{x^{3}}{3}+ln(xlnx+2)$ bạn tự thay cận vào nha