Answer
$$x \ln \left(x^{2}+9\right)-2 x+6 \tan ^{-1}\left(\frac{x}{3}\right)+C$$
Work Step by Step
Given $$\int \ln \left(x^{2}+9\right) d x$$
Let
\begin{align*}
u&= \ln \left(x^{2}+9\right)\ \ \ \ \ \ \ \ \ \ \ \ dv=dx\\
du&=\frac{2x}{x^2+9} \ \ \ \ \ \ \ \ \ \ \ \ \ v=x
\end{align*}
Then
\begin{align*}
\int \ln \left(x^{2}+9\right) d x&=
x \ln \left(x^{2}+9\right)-2 \int \frac{x^{2}}{x^{2}+9} d x\\
&=x \ln \left(x^{2}+9\right)-2 \int\left[1-\frac{9}{x^{2}+9}\right]\\
&= x \ln \left(x^{2}+9\right)-2 x+6 \tan ^{-1}\left(\frac{x}{3}\right)+C
\end{align*}
