Use the substitution $x=3\sin u$ to show that

$${\int }_0^{1.5}\frac{2x+3}{(9-x^2{)}^{\frac{3}{2}}}dx={\int }_0^p\left(\frac{2}{3}\sec u\tan u+\frac{1}{3}{\sec }^{2}u\right)du$$

where $p$ is a constant to be found.

Hence find the exact value of

$${\int }_0^{1.5}\frac{2x+3}{(9-x^2{)}^{\frac{3}{2}}}dx$$