int sqrt{3-2x-x^2} , dx = . . . . . . + C. | vedclass

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(D) To solve the integral $I = \int \sqrt{3-2x-x^2} \, dx$,first complete the square inside the square root: $3-2x-x^2 = 4 - (x^2+2x+1) = 2^2 - (x+1)^

Vedclass · Accepted Answer

$\frac{1}{2}(x+1) \sqrt{3-2x-x^2} + 2 \sin^{-1}\left(\frac{x+1}{2}\right)$

Vedclass · Answer

(D) To solve the integral $I = \int \sqrt{3-2x-x^2} \, dx$,first complete the square inside the square root: $3-2x-x^2 = 4 - (x^2+2x+1) = 2^2 - (x+1)^2$. Now,the integral becomes $I = \int \sqrt{2^2 - (x+1)^2} \, dx$. Using the standard formula $\int \sqrt{a^2 - u^2} \, du = \frac{u}{2} \sqrt{a^2 - u^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{u}{a}\right) + C$,where $u = x+1$ and $a = 2$: $I = \frac{x+1}{2} \sqrt{2^2 - (x+1)^2} + \frac{2^2}{2} \sin^{-1}\left(\frac{x+1}{2}\right) + C$. Simplifying this,we get: $I = \frac{1}{2}(x+1) \sqrt{3-2x-x^2} + 2 \sin^{-1}\left(\frac{x+1}{2}\right) + C$. Thus,the correct option is $D$.

$\int \sqrt{3-2x-x^2} \, dx = $ . . . . . . $+ C$.

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