Đáp án:
$I=\dfrac{3}{16}e^4+\dfrac{1}{16}.$
Giải thích các bước giải:
$I=\displaystyle\int\limits^e_1 x^3\ln x \, dx\\ u=\ln x \Rightarrow du=\dfrac{1}{x} dx\\ dv=x^3 \Rightarrow v=\dfrac{x^4}{4}\\ I= \dfrac{1}{4}x^4\ln x \Bigg\vert^e_1-\dfrac{1}{4}\displaystyle\int\limits^e_1 x^3 \, dx\\ =\dfrac{1}{4}e^4-\dfrac{1}{16}x^4\Bigg\vert^e_1\\ =\dfrac{1}{4}e^4-\dfrac{1}{16}e^4+\dfrac{1}{16}\\ =\dfrac{3}{16}e^4+\dfrac{1}{16}.$
