Answer
$$\int\frac{xdx}{\sqrt{x-2}}=\frac{2\sqrt{x-2}}{3}(x+4)+C$$
Work Step by Step
$$A=\int\frac{xdx}{\sqrt{x-2}}=\int x(x-2)^{-1/2}dx$$
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 $a=1$, $b=-2$ and $n=-1/2$ here, we have
$$A=\frac{(x-2)^{1/2}}{1^2}\Big[\frac{x-2}{\frac{3}{2}}-\frac{(-2)}{\frac{1}{2}}\Big]+C$$ $$A=\sqrt{x-2}\Big[\frac{2(x-2)}{3}+4\Big]+C$$ $$A=\frac{\sqrt{x-2}}{3}(2x-4+12)+C$$ $$A=\frac{\sqrt{x-2}}{3}(2x+8)+C$$ $$A=\frac{2\sqrt{x-2}}{3}(x+4)+C$$
