int frac{1}{sqrt{8+2x-x^2}} dx = | vedclass

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(D) Let $I = \int \frac{dx}{\sqrt{8+2x-x^2}}$.First,complete the square for the quadratic expression inside the square root:$8+2x-x^2 = -(x^2-2x-8) =

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$\sin^{-1}\left(\frac{x-1}{3}\right)+c$

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(D) Let $I = \int \frac{dx}{\sqrt{8+2x-x^2}}$.First,complete the square for the quadratic expression inside the square root:$8+2x-x^2 = -(x^2-2x-8) = -(x^2-2x+1-9) = -( (x-1)^2 - 9 ) = 9 - (x-1)^2$.Now,substitute this back into the integral:$I = \int \frac{dx}{\sqrt{9-(x-1)^2}} = \int \frac{dx}{\sqrt{3^2-(x-1)^2}}$.Using the standard integral formula $\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\frac{x}{a}\right)+c$,we get:$I = \sin^{-1}\left(\frac{x-1}{3}\right)+c$.

$\int \frac{1}{\sqrt{8+2x-x^2}} dx =$

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