If the solve the problem

$\sqrt{2 \mathrm{x}-1}=\mathrm{t} \Rightarrow 2 \mathrm{x}-1=\mathrm{t}^{2} \Rightarrow 2 \mathrm{dx}=2 \mathrm{t} \cdot \mathrm{dt}$

$\int \frac{x+1}{\sqrt{2 x-1}} d x=\int \frac{\frac{t^{2}+1}{2}+1}{t} t d t=\int \frac{t^{2}+3}{2} d t$

$=\frac{1}{2}\left(\frac{t^{3}}{3}+3 t\right)=\frac{t}{6}\left(t^{2}+9\right)+c$

$=\sqrt{2 x-1}\left(\frac{2 x-1+9}{6}\right)+c=\sqrt{2 x-1}\left(\frac{x+4}{3}\right)+c$

$\Rightarrow f(x)=\frac{x+4}{3}$