Answer
$$\frac{1}{2}x^2 \ln (x+12)-\frac{1}{2} \left(\frac{1}{2}(x-12)^2+144\ln|x+12|\right)+C$$
Work Step by Step
Given $$\int x \ln (x+12) d x$$
Let
\begin{align*}
u&=\ln (x+12)\ \ \ \ \ \ \ \ \ \ dv= xdx\\
du&=\frac{1}{x+12}dx\ \ \ \ \ \ \ \ v=\frac{1}{2}x^2
\end{align*}
Then
\begin{align*}
\int x \ln (x+12) d x&=\frac{1}{2}x^2 \ln (x+12)-\frac{1}{2}\int \frac{x^2}{x+12}dx\\
&= \frac{1}{2}x^2 \ln (x+12)-\frac{1}{2}\int \left((x-12)+\frac{144}{x+12}\right) dx\\
&= \frac{1}{2}x^2 \ln (x+12)-\frac{1}{2} \left(\frac{1}{2}(x-12)^2+144\ln|x+12|\right)+C
\end{align*}
