If int frac{x+1}{sqrt{2x-1}} , dx = f(x) sqrt | vedclass

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(C) Let $I = \int \frac{x+1}{\sqrt{2x-1}} \, dx$. Substitute $2x-1 = t$,which implies $x = \frac{t+1}{2}$ and $dx = \frac{1}{2} dt$. $I = \int \frac{\

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$\frac{1}{3}(x+4)$

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(C) Let $I = \int \frac{x+1}{\sqrt{2x-1}} \, dx$. Substitute $2x-1 = t$,which implies $x = \frac{t+1}{2}$ and $dx = \frac{1}{2} dt$. $I = \int \frac{\frac{t+1}{2} + 1}{\sqrt{t}} \cdot \frac{1}{2} \, dt$ $I = \frac{1}{2} \int \frac{\frac{t+3}{2}}{\sqrt{t}} \, dt = \frac{1}{4} \int \left( \frac{t}{\sqrt{t}} + \frac{3}{\sqrt{t}} \right) \, dt$ $I = \frac{1}{4} \int (t^{1/2} + 3t^{-1/2}) \, dt$ $I = \frac{1}{4} \left( \frac{t^{3/2}}{3/2} + 3 \cdot \frac{t^{1/2}}{1/2} \right) + C$ $I = \frac{1}{4} \left( \frac{2}{3} t^{3/2} + 6 t^{1/2} \right) + C$ $I = \left( \frac{1}{6} t^{3/2} + \frac{3}{2} t^{1/2} \right) + C$ $I = \frac{1}{6} t^{1/2} (t + 9) + C$ Substituting $t = 2x-1$ back: $I = \frac{1}{6} \sqrt{2x-1} (2x-1+9) + C$ $I = \frac{1}{6} \sqrt{2x-1} (2x+8) + C$ $I = \frac{1}{6} \cdot 2(x+4) \sqrt{2x-1} + C = \frac{1}{3}(x+4) \sqrt{2x-1} + C$ Comparing with $f(x) \sqrt{2x-1} + C$,we get $f(x) = \frac{1}{3}(x+4)$.

If $\int \frac{x+1}{\sqrt{2x-1}} \, dx = f(x) \sqrt{2x-1} + C$,where $C$ is an arbitrary constant,then $f(x)$ is equal to

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