If int {frac{{x + 1}}{{sqrt {2x - 1} }}} dx = | vedclass

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(D) Let $2x - 1 = t^2$. Then $2 dx = 2t dt$,so $dx = t dt$.Also,$x = \frac{t^2 + 1}{2}$,which implies $x + 1 = \frac{t^2 + 1}{2} + 1 = \frac{t^2 + 3}{

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

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(D) Let $2x - 1 = t^2$. Then $2 dx = 2t dt$,so $dx = t dt$.Also,$x = \frac{t^2 + 1}{2}$,which implies $x + 1 = \frac{t^2 + 1}{2} + 1 = \frac{t^2 + 3}{2}$.Substituting these into the integral:$\int \frac{x + 1}{\sqrt{2x - 1}} dx = \int \frac{(\frac{t^2 + 3}{2})}{t} (t dt) = \int \frac{t^2 + 3}{2} dt$$= \frac{1}{2} (\frac{t^3}{3} + 3t) + C = \frac{t^3}{6} + \frac{3t}{2} + C$$= t (\frac{t^2}{6} + \frac{3}{2}) + C = t (\frac{t^2 + 9}{6}) + C$Since $t = \sqrt{2x - 1}$,we have $t^2 = 2x - 1$.Substituting $t^2$ back:$= \sqrt{2x - 1} (\frac{2x - 1 + 9}{6}) + C = \sqrt{2x - 1} (\frac{2x + 8}{6}) + C$$= \sqrt{2x - 1} (\frac{x + 4}{3}) + C$.Comparing this 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 a constant of integration,then $f(x)$ is equal to

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