Answer
$$\frac{1}{2}\ln2.$$
Work Step by Step
We have
$$\int_0^3 \frac{xdx}{x^2+9}=\frac{1}{2}\int_0^3\frac{2xdx}{x^2+9}=\frac{1}{2}\ln (x^2+9)|_0^3\\
=\frac{1}{2}\ln (x^2+9)|_0^3=\frac{1}{2}(\ln 18-\ln 9)= \frac{1}{2}\ln \frac{18}{9}=\frac{1}{2}\ln2.$$
