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

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(A) Let $t = \sqrt{x - 2}$, then $t^2 = x - 2$, which implies $x = t^2 + 2$.Differentiating both sides, $dx = 2t dt$.Substituting these into the integ

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $t = \sqrt{x - 2}$, then $t^2 = x - 2$, which implies $x = t^2 + 2$.Differentiating both sides, $dx = 2t dt$.Substituting these into the integral:$\int \frac{t}{t^2 + 2} (2t dt) = 2 \int \frac{t^2}{t^2 + 2} dt$$= 2 \int \frac{t^2 + 2 - 2}{t^2 + 2} dt = 2 \int \left( 1 - \frac{2}{t^2 + 2} ight) dt$$= 2 \left( \int 1 dt - 2 \int \frac{1}{t^2 + (\sqrt{2})^2} dt ight)$$= 2 \left( t - 2 \cdot \frac{1}{\sqrt{2}} 	an^{-1} \left( \frac{t}{\sqrt{2}} ight) ight) + c$$= 2t - 2\sqrt{2} 	an^{-1} \left( \frac{t}{\sqrt{2}} ight) + c$Substituting $t = \sqrt{x - 2}$ back, we get:$= 2\sqrt{x - 2} - 2\sqrt{2} 	an^{-1} \left( \frac{\sqrt{x - 2}}{\sqrt{2}} ight) + c$

$\int \frac{\sqrt{x - 2}}{x} dx =$

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