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

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(B) Let $I = \int \frac{d x}{\sqrt{x}(x+9)}$.Substitute $x = t^2$,which implies $d x = 2t \, dt$.Substituting these into the integral:$I = \int \frac{

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

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(B) Let $I = \int \frac{d x}{\sqrt{x}(x+9)}$.Substitute $x = t^2$,which implies $d x = 2t \, dt$.Substituting these into the integral:$I = \int \frac{2t \, dt}{t(t^2 + 9)} = \int \frac{2 \, dt}{t^2 + 3^2}$.Using the standard integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$:$I = 2 \cdot \frac{1}{3} 	an^{-1}(\frac{t}{3}) + C$.Substituting $t = \sqrt{x}$ back into the expression:$I = \frac{2}{3} 	an^{-1}(\frac{\sqrt{x}}{3}) + C$.

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

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