int frac{dx}{sqrt{x}(x+9)} is equal to | vedclass

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(B) Let $I = \int \frac{dx}{\sqrt{x}(x+9)}$.Substitute $x = t^2$,then $dx = 2t \, dt$.The integral becomes:$I = \int \frac{2t \, dt}{t(t^2 + 9)} = \in

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

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(B) Let $I = \int \frac{dx}{\sqrt{x}(x+9)}$.Substitute $x = t^2$,then $dx = 2t \, dt$.The integral becomes:$I = \int \frac{2t \, dt}{t(t^2 + 9)} = \int \frac{2 \, dt}{t^2 + 3^2}$.Using the standard integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1} \left(\frac{x}{a}ight) + C$,we get:$I = 2 \cdot \frac{1}{3} 	an^{-1} \left(\frac{t}{3}ight) + C$.Substituting $t = \sqrt{x}$ back,we obtain:$I = \frac{2}{3} 	an^{-1} \left(\frac{\sqrt{x}}{3}ight) + C$.

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

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