1. ## Integration

2. $\displaystyle \int {\frac{{4x}} {{\sqrt {2x - 1} }}} dx = 2\int {\frac{{2x + 1 - 1}} {{\sqrt {2x - 1} }}} dx = 2\int {\sqrt {2x - 1} + 2\int {\frac{1} {{\sqrt {2x - 1} }}} } dx$

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$

$\displaystyle = \sqrt {2x - 1} \left( {\frac{4} {3}x - \frac{2} {3} + 2} \right) + C = \frac{4} {3}\sqrt {2x - 1} (x - 1) + C$

3. 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:

$\displaystyle \int {\frac{{4x}} {{\sqrt {2x - 1} }}\,dx} = 2\sqrt {2x - 1} \left( {\frac{{2x - 1}} {3} + 1} \right) = \frac{4} {3}(x + 1)\sqrt {2x - 1} + k\quad\blacksquare$

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Peritus, I'd re-check your last conclusion.

4. Peritus, I'd re-check your last conclusion
it's not even a conclusion it'ss pure bull@#@. thanks for pointing out my mistake. I've corrected my post.