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Integrate the function
$\sqrt{x^2+4 x+6}$
$\sqrt{x^2+4 x+6}$
Solution:
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Verified Answer
$\int \sqrt{x^2+4 x+6} d x=\int \sqrt{(x+2)^2+(\sqrt{2})^2} d x$
$=\frac{x+2}{2} \sqrt{x^2+4 x+6}+\log \left|(x+2)+\sqrt{x^2+4 x+6}\right|+C$
$=\frac{x+2}{2} \sqrt{x^2+4 x+6}+\log \left|(x+2)+\sqrt{x^2+4 x+6}\right|+C$
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