find the following integral:

$\displaystyle \int^3_0 \frac{x^3}{(3+x^2)^{\frac{5}{2}}} dx$

what i was thinking was:

$\displaystyle x = \sqrt{3}tan u$

$\displaystyle dx = \sqrt{3}sec˛u du$

then i get:

$\displaystyle \int^3_0 \frac{(\sqrt{3}tanu)^3}{(3+(\sqrt{3}tanu)^2)^{\fra c{5}{2}}} \cdot \sqrt{3} sec^2u du$

is this right?