Let $I=\int(1+x) \log x d x$
Put $\quad \log x=t \quad \Rightarrow x=e^{t} \Rightarrow d x=e^{t} d t$
$\begin{aligned}
\therefore I &=\int\left(1+e^{t}\right) t \cdot e^{t} d t=\int\left(t e^{t}+t e^{2 t}\right) d t \\
&=\int t e^{t} d t+\int t e^{2 t} d t=\left[t e^{t}-\int e^{t} d t\right]+\left[t \frac{e^{2 t}}{2}-\int \frac{e^{2 t}}{2} d t\right] \\
&=t e^{t}-e^{t}+t \frac{e^{2 t}}{2}-\frac{e^{2 t}}{4}+C=x \log x-x+\frac{1}{2}(\log x) \cdot x^{2}-\frac{x^{2}}{4}+C \\
&=\left(x+\frac{x^{2}}{2}\right) \log x-\left(x+\frac{x^{2}}{4}\right)+C
\end{aligned}$