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$\int \sqrt{5-2 x+x^2} d x$ is equals to
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Verified Answer
The correct answer is:
$\begin{aligned} \frac{x-1}{2} \sqrt{5-2 x+x^2}+2 \log \mid(x-1) \\ +\sqrt{5-2 x+x^2} \mid+C\end{aligned}$
Let $\begin{aligned} I & =\int \sqrt{5-2 x+x^2} d x \\ I & =\int \sqrt{(x-1)^2+(2)^2}\end{aligned}$
We know that
$\begin{aligned} & {\left[\because \int \sqrt{x^2+a^2}=\frac{x}{2} \sqrt{x^2+a^2}\right.} \\ & \left.\quad+\frac{a^2}{2} \log \left|x+\sqrt{x^2+a^2}\right|\right] \\ & =\frac{x-1}{2} \sqrt{5-2 x+x^2} \\ & \quad+2 \log \left|(x-1)+\sqrt{5-2 x+x^2}\right|+C\end{aligned}$
We know that
$\begin{aligned} & {\left[\because \int \sqrt{x^2+a^2}=\frac{x}{2} \sqrt{x^2+a^2}\right.} \\ & \left.\quad+\frac{a^2}{2} \log \left|x+\sqrt{x^2+a^2}\right|\right] \\ & =\frac{x-1}{2} \sqrt{5-2 x+x^2} \\ & \quad+2 \log \left|(x-1)+\sqrt{5-2 x+x^2}\right|+C\end{aligned}$
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