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

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

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

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(B) Let $I = \int \frac{x+1}{\sqrt{2x-1}} \, dx$. Substitute $2x-1 = t^2$,which implies $x = \frac{t^2+1}{2}$ and $dx = t \, dt$. Then $x+1 = \frac{t^2+1}{2} + 1 = \frac{t^2+3}{2}$. Substituting these into the integral: $I = \int \frac{(\frac{t^2+3}{2}) t \, dt}{t} = \int \frac{t^2+3}{2} \, dt$. $I = \frac{1}{2} (\frac{t^3}{3} + 3t) + c = \frac{t^3}{6} + \frac{3t}{2} + c$. Factor out $\frac{t}{6}$: $I = \frac{t}{6} (t^2 + 9) + c$. Substitute $t = \sqrt{2x-1}$ back: $I = \frac{\sqrt{2x-1}}{6} (2x-1+9) + c = \frac{\sqrt{2x-1}}{6} (2x+8) + c$. $I = \sqrt{2x-1} \cdot \frac{2(x+4)}{6} + c = \sqrt{2x-1} \cdot \frac{x+4}{3} + c$. Comparing this with $f(x) \sqrt{2x-1} + c$,we get $f(x) = \frac{x+4}{3}$.

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

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