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