Answer
$$\frac{32\ln \left(2\right)}{3}-\frac{7}{4}$$
Work Step by Step
Given $$ \int_{1}^{2}x^5\ln xdx$$
Let
\begin{align*}
u&=\ln x\ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=x^5\\
u&=\frac{1}{x}dx\ \ \ \ \ \ \ \ \ \ \ \ \ dv= \frac{1}{6}x^6
\end{align*}
Then using integration by parts
\begin{align*}
\int_{1}^{2}x^5\ln xdx&=uv-\int vdu\\
&= \frac{1}{6}x^6\ln x\bigg|_{1}^{2}- \frac{1}{6}\int_{1}^{2} x^5\\
&=\frac{1}{6}x^6\ln x\bigg|_{1}^{2}- \frac{1}{36}x^6\bigg|_{1}^{2}\\
&=\frac{32}{3}\ln \left(2\right)-\frac{16}{9}+ \frac{1}{36} \\
&=\frac{32\ln \left(2\right)}{3}-\frac{7}{4}
\end{align*}
