Answer
$$\frac{1}{5}$$
Work Step by Step
$$\eqalign{
& \int_2^{\sqrt 5 } {x{{\left( {{x^2} - 4} \right)}^{3/2}}dx} \cr
& {\text{Let }}u = {x^2} - 4,{\text{ }}du = 2xdx \cr
& {\text{The new limits of integration are}} \cr
& x = \sqrt 5 \to u = 1 \cr
& x = 2 \to u = 0 \cr
& {\text{Substituting}} \cr
& \int_2^{\sqrt 5 } {x{{\left( {{x^2} - 4} \right)}^{3/2}}dx} = \int_0^1 {{u^{3/2}}\left( {\frac{1}{2}} \right)du} \cr
& = \frac{1}{2}\int_0^1 {{u^{3/2}}du} \cr
& {\text{Integrating}} \cr
& = \frac{1}{2}\left[ {\frac{{{u^{5/2}}}}{{5/2}}} \right]_0^1 \cr
& = \frac{1}{5}\left[ {{u^{5/2}}} \right]_0^1 \cr
& = \frac{1}{5}\left( {1 - 0} \right) \cr
& = \frac{1}{5} \cr} $$
