Answer
$$\frac{6}{5}$$
Work Step by Step
$$\eqalign{
& \int_{3/2}^2 {2x\sqrt {2x - 3} dx} \cr
& {\text{Let }}u = 2x - 3,{\text{ }}x = \frac{{u + 3}}{2},{\text{ }}dx = \frac{1}{2}du \cr
& {\text{ }}u = 2x - 3.{\text{ The new limits of integration are}} \cr
& x = 2 \Rightarrow {\text{ }}u = 2\left( 2 \right) - 3 = 1 \cr
& x = 3/2 \Rightarrow {\text{ }}u = 2\left( {3/2} \right) - 3 = 0 \cr
& {\text{Write in terms of }}u \cr
& \int_{3/2}^2 {2x\sqrt {2x - 3} dx} = \int_0^1 {2\left( {\frac{{u + 3}}{2}} \right)\sqrt u \left( {\frac{1}{2}} \right)du} \cr
& = \frac{1}{2}\int_0^1 {\left( {u + 3} \right)\sqrt u du} \cr
& = \frac{1}{2}\int_0^1 {\left( {{u^{3/2}} + 3{u^{1/2}}} \right)du} \cr
& {\text{Integrate by the power rule }}\int {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
& = \frac{1}{2}\left( {\frac{{{u^{5/2}}}}{{5/2}} + 3\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right)} \right)_0^1 \cr
& = \frac{1}{2}\left( {\frac{{2{u^{5/2}}}}{5} + 2{u^{3/2}}} \right)_0^1 \cr
& {\text{Evaluating}} \cr
& = \frac{1}{2}\left( {\frac{{2{{\left( 1 \right)}^{5/2}}}}{5} + 2{{\left( 1 \right)}^{3/2}}} \right) - \frac{1}{2}\left( {\frac{{2{{\left( 0 \right)}^{5/2}}}}{5} + 2{{\left( 0 \right)}^{3/2}}} \right) \cr
& = \frac{1}{2}\left( {\frac{2}{5} + 2} \right) - \frac{1}{2}\left( 0 \right) \cr
& = \frac{6}{5} \cr} $$
