1. We need to evaluate I=int \frac{x+1}{\sqrt{2x-1}}\,dx and write it in the form $I=f(x)\sqrt{2x-1}+C.$

2. Use substitution: Let $u=2x-1 \implies du=2\,dx, \quad dx=\frac{du}{2}, \quad x=\frac{u+1}{2}.$ Then $x+1=\frac{u+1}{2}+1=\frac{u+3}{2}.$

So the integral becomes $I=\int \frac{(u+3)/2}{\sqrt{u}}\cdot \frac{du}{2}$ $=\frac14\int \frac{u+3}{\sqrt{u}}\,du$ $=\frac14\int \left(u^{1/2}+3u^{-1/2}\right)du.$

3. Integrate term by term: $I=\frac14\left(\int u^{1/2}du+3\int u^{-1/2}du\right)$ $=\frac14\left(\frac{2}{3}u^{3/2}+3\cdot 2u^{1/2}\right)+C$ $=\frac{1}{6}u^{3/2}+\frac{3}{2}u^{1/2}+C.$

4. Substitute back $u=2x-1$: $I=\frac{1}{6}(2x-1)^{3/2}+\frac{3}{2}(2x-1)^{1/2}+C.$ Factor out $\sqrt{2x-1}$: $I=\sqrt{2x-1}\left(\frac{1}{6}(2x-1)+\frac{3}{2}\right)+C.$

Simplify inside the bracket: $\frac{1}{6}(2x-1)+\frac{3}{2}=\frac{2x-1}{6}+\frac{9}{6}=\frac{2x+8}{6}=\frac{x+4}{3}.$

Hence, $I=\frac{x+4}{3}\sqrt{2x-1}+C.$ So, $f(x)=\frac{1}{3}(x+4).$

5. Check options:

• A: $\frac{2}{3}(x-4)$  incorrect
• B: $\frac{1}{3}(x+4)$  correct
• C: $\frac{1}{3}(x+1)$  incorrect
• D: $\frac{2}{3}(x+2)$  incorrect

Therefore, the correct option is B.