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