Answer
$$\int_{1}^{e} x^3 \ln x d x
=\frac{3e^{4}+1 }{16} $$
Work Step by Step
Given $$\int_{1}^{e} x^3 \ln x d x $$ Now, using integration by parts:
$$ u=\ln x \Rightarrow du= \frac{1}{x}dx $$ $$ dv=x^3 dx \Rightarrow v= \frac{x^4}{4} $$ So, we get
\begin{aligned}
I&=\int_{1}^{e} x^4 \ln x d x\\
&=uv|_{1}^{e}- \int_{1}^{e}vdu\\
&=\frac{x^{4} \ln x}{4}|_{1}^{e}-\frac{1}{4} \int_{1}^{e} \frac{x^{4}}{x} d x\\
&=\frac{e^{4} \ln e}{4}-\frac{1^{2} \ln1}{4}-\frac{1}{4} \int_{1}^{e}x^3d x\\
&=\frac{e^{4} \ }{4}-\frac{x^4}{16} |_{1}^{e} \\
&=\frac{e^{4} }{4}-\frac{e^4}{16}+ \frac{1}{16} \\ \\
&=\frac{4e^{4} }{16}-\frac{e^4}{16}+ \frac{1}{16} \\ \\
&=\frac{3e^{4}+1 }{16} \\ \\
\end{aligned}
