Answer
$\displaystyle \frac{1}{2}(1+x)(x-1)\ln(1+x)-\frac{1}{4}(1+x)^{2}+1+x+C$
Work Step by Step
Substitute$ \left[\begin{array}{ll}
t=1+x, & x=t-1\\
dt=dx &
\end{array}\right]$
$I=\displaystyle \int x\ln(1+x)dx=\int(t-1)\ln t$
By parts, $\displaystyle \left[\begin{array}{ll}
u=\ln t & dv=(t-1)\\
& \\
du=\frac{dt}{t} & v=\frac{1}{2}t^{2}-t
\end{array}\right],\quad\int udv=uv-\int vdu$
$I=(\displaystyle \frac{1}{2}t^{2}-t)\ln t-\int(\frac{1}{2}t^{2}-t)\cdot\frac{1}{t}dt$
$=(\displaystyle \frac{1}{2}t^{2}-t)\ln t-\int(\frac{1}{2}t-1)dt$
$=\displaystyle \frac{1}{2}t(t-2)\ln t-(\frac{1}{2}\cdot\frac{t^{2}}{2}-t)+C$
... bring x back ...($t=1+x$)
$I=\displaystyle \frac{1}{2}(1+x)(x-1)\ln(1+x)-\frac{1}{4}(1+x)^{2}+1+x+C$
