$$
\text { } \begin{aligned}
I & =\int_0^1(1+x) \log (1+x) d x \\
& =\left[(\log (1+x))\left(x+\frac{x^2}{2}\right)\right]_0^1-\int_0^1 \frac{\left(x+\frac{x^2}{2}\right)}{1+x} d x \\
& =\frac{3}{2} \log _e 2-\frac{1}{2} \int_0^1\left[(x+1)-\frac{1}{1+x}\right] d x \\
& =\frac{3}{2} \log _e 2-\frac{1}{2}\left[\frac{x^2}{2}+x-\log _e(1+x)\right]_0^1
\end{aligned}
$$


$$
\begin{aligned}
& =\frac{3}{2} \log _e 2-\frac{1}{2}\left[\frac{1}{2}+1-\log 2\right] \\
& =2\left(\log _e 2\right)-\frac{3}{4}=-\frac{3}{4}+2 \log _e 2
\end{aligned}
$$