Answer
$$\frac{1}{2}\left(\left(x^{2}+1\right) \ln \left(x^{2}+1\right)-\left(x^{2}+1\right)\right)+c$$
Work Step by Step
Given: $$\int x \ln \left(x^{2}+1\right) d x$$
Let $u=x^2+1\ \to\ \ du=2xdx$, so:
\begin{align*}
\int x \ln \left(x^{2}+1\right) d x&=\frac{1}{2}\int \ln \left(u\right) du
\end{align*}
Apply the Natural Log Rule for antiderivatives:
$$
\int \ln u d u=u \ln u-u+c
$$
Thus:
\begin{align*}
\int x \ln \left(x^{2}+1\right) d x&=\frac{1}{2}\int \ln \left(u\right) du \\
&=\frac{1}{2}\left(u \ln u-u+\right)c\\
&=\frac{1}{2}\left(\left(x^{2}+1\right) \ln \left(x^{2}+1\right)-\left(x^{2}+1\right)\right)+c
\end{align*}