Answer
$$\int\frac{xdx}{(2x+3)^{3/2}}=\frac{x+3}{\sqrt{2x+3}}+C$$
Work Step by Step
$$A=\int\frac{xdx}{(2x+3)^{3/2}}=\int x(2x+3)^{-3/2}dx$$
Using Formula 22, which states that
$$\int x(ax+b)^ndx=\frac{(ax+b)^{n+1}}{a^2}\Big[\frac{ax+b}{n+2}-\frac{b}{n+1}\Big]+C$$
with $a=2$, $b=3$ and $n=-3/2$ here, we have
$$A=\frac{(2x+3)^{-1/2}}{2^2}\Big[\frac{2x+3}{\frac{1}{2}}-\frac{3}{-\frac{1}{2}}\Big]+C$$ $$A=\frac{1}{4\sqrt{2x+3}}\Big[4x+6+6\Big]+C$$ $$A=\frac{4x+12}{4\sqrt{2x+3}}+C$$ $$A=\frac{x+3}{\sqrt{2x+3}}+C$$
