Official solution
\[\sqrt {2x - 1} = t \Rightarrow 2x - 1 = {t^2} \Rightarrow 2dx = 2t.dt\]
\[\int {{{x + 1} \over {\sqrt {2x - 1} }}dx = \int {{{{{{t^2} + 1} \over 2} + 1} \over t}tdt = \int {{{{t^2} + 3} \over 2}dt} } } \]
\[ = {1 \over 2}\left( {{{{t^3}} \over 3} + 3t} \right) = {t \over 6}\left( {{t^2} + 9} \right) + c\]
\[ = \sqrt {2x - 1} \left( {{{2x - 1 + 9} \over 6}} \right) + c = \sqrt {2x - 1} \left( {{{x + 4} \over 3}} \right) + c\]
\[ \Rightarrow f\left( x \right) = {{x + 4} \over 3}\]