Answer
$$6$$
Work Step by Step
$$\eqalign{
& \int_1^3 {x\root 3 \of {{x^2} - 1} } dx \cr
& {\text{set }}{u^3} = {x^2} - 1,\,\,\,\,\,{\text{then }}3{u^2}du = 2xdx,\,\,\,\,\,\,\,\,xdx = \frac{3}{2}{u^2}du \cr
& {\text{switch the limits of integration using }}u = \root 3 \of {{x^2} - 1} \cr
& x = 1 \to u = 0 \cr
& x = 3 \to u = 2 \cr
& {\text{use the change of variable}} \cr
& \int_1^3 {x\root 3 \of {{x^2} - 1} } dx = \int_0^2 {\root 3 \of {{u^3}} \left( {\frac{3}{2}{u^2}du} \right)} \cr
& = \frac{3}{2}\int_0^2 {{u^3}du} \cr
& {\text{integrate}} \cr
& = \frac{3}{2}\left( {\frac{{{u^4}}}{4}} \right)_0^2 \cr
& {\text{evaluate the limits}} \cr
& = \frac{3}{2}\left( {\frac{{{{\left( 2 \right)}^4}}}{4} - 0} \right) \cr
& = 6 \cr} $$
