The value of int frac{dx}{sqrt{x}(x + 9)} is | vedclass

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(D) Let $I = \int \frac{dx}{\sqrt{x}(x + 9)}$.Substitute $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t \, dt$.Substituting these into the inte

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$\frac{2}{3}	an^{-1}\left(\frac{\sqrt{x}}{3}ight)$

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(D) Let $I = \int \frac{dx}{\sqrt{x}(x + 9)}$.Substitute $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t \, dt$.Substituting these into the integral,we get:$I = \int \frac{2t \, dt}{t(t^2 + 9)} = 2 \int \frac{dt}{t^2 + 3^2}$.Using the standard integral formula $\int \frac{du}{u^2 + a^2} = \frac{1}{a} 	an^{-1}\left(\frac{u}{a}ight) + C$,we have:$I = 2 \cdot \frac{1}{3} 	an^{-1}\left(\frac{t}{3}ight) + C = \frac{2}{3} 	an^{-1}\left(\frac{\sqrt{x}}{3}ight) + C$.

The value of $\int \frac{dx}{\sqrt{x}(x + 9)}$ is equal to

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