Answer
$8\sqrt 3$
Work Step by Step
$u$ = $x-3$
$du$ = $dx$
$\int{\frac{x}{\sqrt {x-3}}}dx$ = $\int{\frac{u+3}{\sqrt u}}du$ = $\frac{2}{3}u^{\frac{3}{2}}+6u^{\frac{1}{2}}+C$ = $\frac{2}{3}(x-3)^{\frac{3}{2}}+6(x-3)^{\frac{1}{2}}+C$
$\int_3^6{\frac{x}{\sqrt {x-3}}}dx$ = $\lim\limits_{R \to 3^{+}}$$\int_R^6{\frac{x}{\sqrt {x-3}}}dx$ = $\lim\limits_{R \to 3^{+}}[8\sqrt 3-\frac{2}{3}(R-3)^{\frac{3}{2}}-6(R-3)^{\frac{1}{2}}]$ = $8\sqrt 3$
