Answer
$$
\int t \ln (t+1) d t= \frac{1}{2}t\ln (t+1)-\frac{1}{2}\left(\frac{1}{2}(t-1)^2+ \ln(t+1)\right)+c
$$
Work Step by Step
$$
\int t \ln (t+1) d t
$$
Integrate by parts
\begin{align*}
u&=\ln (t+1)\ \ \ \ \ \ dv=tdt\\
du&=\frac{1}{t+1}\ \ \ \ \ \ v=\frac{1}{2}t^2
\end{align*}
Then
\begin{align*}
\int t \ln (t+1) d t&= \frac{1}{2}t\ln (t+1)-\frac{1}{2}\int \frac{t^2}{t+1}dt\\
&= \frac{1}{2}t\ln (t+1)-\frac{1}{2}\int\left((t-1)+ \frac{1}{t+1}\right)dt\\
&= \frac{1}{2}t\ln (t+1)-\frac{1}{2}\left(\frac{1}{2}(t-1)^2+ \ln(t+1)\right)+c
\end{align*}
