int frac{dx}{sqrt{7-6x-x^2}} is equal to | vedclass

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(C) Let $I = \int \frac{dx}{\sqrt{7-6x-x^2}}$.First,complete the square for the quadratic expression $7-6x-x^2$:$7-6x-x^2 = 7 - (x^2 + 6x) = 7 - (x^2

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$\sin^{-1}\left(\frac{x+3}{4}\right)+C$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{\sqrt{7-6x-x^2}}$.First,complete the square for the quadratic expression $7-6x-x^2$:$7-6x-x^2 = 7 - (x^2 + 6x) = 7 - (x^2 + 6x + 9 - 9) = 7 - ((x+3)^2 - 9) = 16 - (x+3)^2 = 4^2 - (x+3)^2$.Now,substitute this into the integral:$I = \int \frac{dx}{\sqrt{4^2 - (x+3)^2}}$.Using the standard integral formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C$,where $a = 4$ and $x$ is replaced by $(x+3)$:$I = \sin^{-1}\left(\frac{x+3}{4}\right) + C$.

$\int \frac{dx}{\sqrt{7-6x-x^2}}$ is equal to

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