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