$\displaystyle
\int {\frac{{4x}}
{{\sqrt {2x - 1} }}} dx = 2\int {\frac{{2x + 1 - 1}}
{{\sqrt {2x - 1} }}} dx = 2\int {\sqrt {2x - 1} + 2\int {\frac{1}
{{\sqrt {2x - 1} }}} } dx
$
and now the result follows immediately:
$\displaystyle
= {\raise0.7ex\hbox{$2$} \!\mathord{\left/
{\vphantom {2 3}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$3$}}\left( {2x - 1} \right)^{{\raise0.7ex\hbox{$3$} \!\mathord{\left/
{\vphantom {3 2}}\right.\kern-\nulldelimiterspace}
\!\lower0.7ex\hbox{$2$}}} + 2\sqrt {2x - 1} + C
$
$\displaystyle
= \sqrt {2x - 1} \left( {\frac{4}
{3}x - \frac{2}
{3} + 2} \right) + C = \frac{4}
{3}\sqrt {2x - 1} (x - 1) + C
$
I'd tackle this as follows:
Substitute $\displaystyle u^2=2x-1\implies u\,du=dx,$ the integral becomes
$\displaystyle 2\int {\left( {u^2 + 1} \right)\,du} = 2\left( {\frac{1}
{3}u^3 + u} \right) + k = 2u\left( {\frac{1}
{3}u^2 + 1} \right) + k.$
Back substitute:
$\displaystyle \int {\frac{{4x}}
{{\sqrt {2x - 1} }}\,dx} = 2\sqrt {2x - 1} \left( {\frac{{2x - 1}}
{3} + 1} \right) = \frac{4}
{3}(x + 1)\sqrt {2x - 1} + k\quad\blacksquare$
---
Peritus, I'd re-check your last conclusion.