int sqrt{5-2x+x^2} dx is equal to | vedclass

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(C) Let $I = \int \sqrt{5-2x+x^2} dx$. Completing the square,we have $5-2x+x^2 = (x-1)^2 + 4 = (x-1)^2 + 2^2$. Using the standard integral formula $\i

Vedclass · Accepted Answer

$\frac{x-1}{2} \sqrt{5-2x+x^2} + 2 \log |(x-1) + \sqrt{5-2x+x^2}| + C$

Vedclass · Answer

(C) Let $I = \int \sqrt{5-2x+x^2} dx$. Completing the square,we have $5-2x+x^2 = (x-1)^2 + 4 = (x-1)^2 + 2^2$. Using the standard integral formula $\int \sqrt{x^2+a^2} dx = \frac{x}{2} \sqrt{x^2+a^2} + \frac{a^2}{2} \log |x + \sqrt{x^2+a^2}| + C$,where $x$ is replaced by $(x-1)$ and $a=2$: $I = \frac{x-1}{2} \sqrt{(x-1)^2 + 2^2} + \frac{2^2}{2} \log |(x-1) + \sqrt{(x-1)^2 + 2^2}| + C$. $I = \frac{x-1}{2} \sqrt{5-2x+x^2} + 2 \log |(x-1) + \sqrt{5-2x+x^2}| + C$.

$\int \sqrt{5-2x+x^2} dx$ is equal to

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