Answer
$6-3\sqrt{3}$
Work Step by Step
We begin with the integral:
$$\int_{0}^{3}\frac{x}{\sqrt{36-x^2}}dx$$
We can now introduce a u-substitution
$$u=36-x^2\\ du=-2xdx\\\frac{-1}{2}du=xdx$$
After substituting and changing the bounds, we get:
$$\frac{-1}{2}\int_{36}^{27}\frac{du}{\sqrt{u}}=\frac{-1}{2}\bigg[2\sqrt{u}\bigg]_{36}^{27}=\frac{-1}{2}(2\sqrt{27}-2\sqrt{36})=6-3\sqrt{3}$$
