I expect this is actually $\displaystyle \displaystyle \int{2x\sqrt{2x-3}\,dx} = \int{\frac{2x(\sqrt{2x-3})^2}{\sqrt{2x-3}}\,dx}$

$\displaystyle \displaystyle = \int{2x(\sqrt{2x-3})^2\left(\frac{1}{\sqrt{2x-3}}\right)\,dx}$.

Now make the substitution $\displaystyle \displaystyle u = \sqrt{2x-3}$.