Answer
$$\int x\sqrt{4x-x^2}dx=\frac{(x+2)(x-3)\sqrt{4x-x^2}}{3}+4\sin^{-1}\Big(\frac{x-2}{2}\Big)+C$$
Work Step by Step
$$A=\int x\sqrt{4x-x^2}dx$$
Use the Formula 124, which states that $$\int x\sqrt{2ax-x^2}dx=\frac{(x+a)(2x-3a)\sqrt{2ax-x^2}}{6}+\frac{a^3}{2}\sin^{-1}\Big(\frac{x-a}{a}\Big)+C$$
for $a=2$ here:
$$A=\frac{(x+2)(2x-6)\sqrt{4x-x^2}}{6}+\frac{8}{2}\sin^{-1}\Big(\frac{x-2}{2}\Big)+C$$ $$A=\frac{(x+2)(x-3)\sqrt{4x-x^2}}{3}+4\sin^{-1}\Big(\frac{x-2}{2}\Big)+C$$
