int frac{dx}{sqrt{9-8x-4x^2}} = . . . . . . | vedclass

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(D) To evaluate the integral $I = \int \frac{dx}{\sqrt{9-8x-4x^2}}$, we first complete the square for the quadratic expression in the denominator.$9 -

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$\frac{1}{2} \sin^{-1} (\frac{2x+2}{\sqrt{13}})$

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(D) To evaluate the integral $I = \int \frac{dx}{\sqrt{9-8x-4x^2}}$, we first complete the square for the quadratic expression in the denominator.$9 - 8x - 4x^2 = 9 - (4x^2 + 8x) = 9 - 4(x^2 + 2x)$.Adding and subtracting $1$ inside the bracket: $9 - 4(x^2 + 2x + 1 - 1) = 9 - 4(x+1)^2 + 4 = 13 - (2x+2)^2$.Thus, the integral becomes $I = \int \frac{dx}{\sqrt{(\sqrt{13})^2 - (2x+2)^2}}$.Using the standard formula $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{a}) + C$, where $u = 2x+2$ and $du = 2dx$ (so $dx = \frac{du}{2}$):$I = \frac{1}{2} \int \frac{du}{\sqrt{(\sqrt{13})^2 - u^2}} = \frac{1}{2} \sin^{-1}(\frac{u}{\sqrt{13}}) + C$.Substituting $u = 2x+2$, we get $I = \frac{1}{2} \sin^{-1}(\frac{2x+2}{\sqrt{13}}) + C$.

$\int \frac{dx}{\sqrt{9-8x-4x^2}} = $ . . . . . . + $C$

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