If \(\int \frac{x+1}{\sqrt{2 x-1}} \mathrm{~d} x=\mathrm{f}(x) \sqrt{2 x-1}+\mathrm{c}\), (where c is a constant of integration), then \(\mathrm{f}(x)\) is equal to

Let \(\mathrm{I}=\int \frac{x+1}{\sqrt{2 x-1}} \mathrm{~d} x\)
Put \(2 x-1=\mathrm{t}^2 \Rightarrow \mathrm{x}+1=\frac{\mathrm{t}^2+3}{2}\)
\(\begin{aligned}
\Rightarrow & d x=t \mathrm{dt} \\
\therefore \quad & =\int \frac{\left(\frac{\mathrm{t}^2+3}{2}\right) \mathrm{tdt}}{\mathrm{t}} \\
& =\frac{\mathrm{t}^3}{6}+\frac{3}{2} \mathrm{t}+\mathrm{c} \\
& =\frac{\mathrm{t}}{2}\left(\frac{\mathrm{t}^2+9}{3}\right)+\mathrm{c} \\
& =\frac{\sqrt{2 x-1}}{2}\left(\frac{2 x-1+9}{3}\right)+\mathrm{c} \\
& =\sqrt{2 x-1}\left(\frac{x+4}{3}\right)+\mathrm{c}
\end{aligned}\)
Comparing with \(\mathrm{f}(x) \sqrt{2 x-1}+\mathrm{c}\), we get
\(\mathrm{f}(x)=\frac{x+4}{3}\)