int frac{sqrt{x-2}}{2x+4} dx= | vedclass

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(B) Let $I = \int \frac{\sqrt{x-2}}{2x+4} dx$. Substitute $t = \sqrt{x-2}$,then $t^2 = x-2$,so $x = t^2+2$ and $dx = 2t dt$. Substituting these into t

Vedclass · Accepted Answer

$\sqrt{x-2}-2 \operatorname{Tan}^{-1}\left(\frac{\sqrt{x-2}}{2}\right)+c$

Vedclass · Answer

(B) Let $I = \int \frac{\sqrt{x-2}}{2x+4} dx$. Substitute $t = \sqrt{x-2}$,then $t^2 = x-2$,so $x = t^2+2$ and $dx = 2t dt$. Substituting these into the integral: $I = \int \frac{t}{2(t^2+2)+4} (2t dt) = \int \frac{2t^2}{2t^2+8} dt = \int \frac{t^2}{t^2+4} dt$. Now,rewrite the integrand: $I = \int \frac{t^2+4-4}{t^2+4} dt = \int \left(1 - \frac{4}{t^2+4}\right) dt$. Integrating term by term: $I = \int 1 dt - 4 \int \frac{1}{t^2+2^2} dt = t - 4 \left(\frac{1}{2} \operatorname{Tan}^{-1}\left(\frac{t}{2}\right)\right) + c$. $I = t - 2 \operatorname{Tan}^{-1}\left(\frac{t}{2}\right) + c$. Substituting $t = \sqrt{x-2}$ back: $I = \sqrt{x-2} - 2 \operatorname{Tan}^{-1}\left(\frac{\sqrt{x-2}}{2}\right) + c$.

$\int \frac{\sqrt{x-2}}{2x+4} dx=$

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