Answer
$\frac{{16}}{3}$
Work Step by Step
$$\eqalign{
& \int_1^5 {\frac{x}{{\sqrt {2x - 1} }}} dx \cr
& {\text{Let }}u = 2x - 1,{\text{ }}x = \frac{1}{2}\left( {u + 1} \right),{\text{ }}dx = \frac{1}{2}du \cr
& {\text{The new limits of integration are:}} \cr
& x = 5 \to u = 9 \cr
& x = 1 \to u = 1 \cr
& \int_1^5 {\frac{x}{{\sqrt {2x - 1} }}} dx = \int_1^9 {\frac{1}{2}\left( {u + 1} \right)\left( {\frac{1}{{\sqrt u }}} \right)\left( {\frac{1}{2}} \right)} du \cr
& = \frac{1}{4}\int_1^9 {\frac{{u + 1}}{{\sqrt u }}} du \cr
& = \frac{1}{4}\int_1^9 {\left( {{u^{1/2}} + {u^{ - 1/2}}} \right)} du \cr
& {\text{Integrating}} \cr
& = \frac{1}{4}\left( {\frac{2}{3}{u^{3/2}} + 2{u^{1/2}}} \right)_1^9 \cr
& = \frac{1}{4}\left( {\frac{2}{3}{{\left( 9 \right)}^{3/2}} + 2{{\left( 9 \right)}^{1/2}}} \right) - \frac{1}{4}\left( {\frac{2}{3}{{\left( 1 \right)}^{3/2}} + 2{{\left( 1 \right)}^{1/2}}} \right) \cr
& {\text{Simplifying}} \cr
& = 6 - \frac{2}{3} \cr
& = \frac{{16}}{3} \cr} $$
