Answer
$$\frac{1}{2}\ln \left(x+1\right)\left(x+1\right)^2-x\ln \left(x+1\right)-\ln \left(x+1\right)-\frac{1}{4}\left(x+1\right)^2+x+1+C$$
Work Step by Step
Given $$\int x\ln (x+1)dx$$
Let $t=x+1\ \ \Rightarrow dt=dx $
$$\int x\ln (x+1)dx=\int (t-1)\ln tdt$$
Let
\begin{align*}
u&=\ln t\ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=t-1\\
u&=\frac{1}{t}dt\ \ \ \ \ \ \ \ \ \ \ \ \ dv= \frac{1}{2}(t-1)^2
\end{align*}
Then using integration by parts
\begin{align*}
\int (t-1)\ln tdt &=uv-\int vdu\\
&= \frac{1}{2}(t-1)^2\ln t- \frac{1}{ 2}\int \frac{(t-1)^2}{ t}dt\\
&= \frac{1}{2}(t-1)^2\ln t- \frac{1}{ 2}\int [t-2+ \frac{1}{ t}]dt\\
&= \frac{1}{2}(t-1)^2\ln t- \frac{1}{ 2}\left(\frac{1}{2}t^2-2t+\ln t\right)+C\\
&=\frac{1}{2}\ln \left(x+1\right)\left(x+1\right)^2-x\ln \left(x+1\right)-\ln \left(x+1\right)-\frac{1}{4}\left(x+1\right)^2+x+1+C
\end{align*}
