int frac{x}{sqrt{x+4}} , dx = . . . . . . | vedclass

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(B) To evaluate the integral $I = \int \frac{x}{\sqrt{x+4}} \, dx$,let $u = x+4$. Then $du = dx$ and $x = u-4$. Substituting these into the integral,w

Vedclass · Accepted Answer

$\frac{2}{3} \sqrt{x+4}(x-8)$

Vedclass · Answer

(B) To evaluate the integral $I = \int \frac{x}{\sqrt{x+4}} \, dx$,let $u = x+4$. Then $du = dx$ and $x = u-4$. Substituting these into the integral,we get: $I = \int \frac{u-4}{\sqrt{u}} \, du$ $I = \int (\frac{u}{\sqrt{u}} - \frac{4}{\sqrt{u}}) \, du$ $I = \int (u^{1/2} - 4u^{-1/2}) \, du$ Integrating term by term: $I = \frac{u^{3/2}}{3/2} - 4 \frac{u^{1/2}}{1/2} + C$ $I = \frac{2}{3} u^{3/2} - 8 u^{1/2} + C$ $I = \frac{2}{3} u^{1/2} (u - 12) + C$ Substituting $u = x+4$ back: $I = \frac{2}{3} \sqrt{x+4} (x+4 - 12) + C$ $I = \frac{2}{3} \sqrt{x+4} (x-8) + C$ Thus,the correct option is $B$.

$\int \frac{x}{\sqrt{x+4}} \, dx = $ . . . . . . $+ C, x > -4$.

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