Integrate the function: int sqrt{x^{2}+4 x+6} | vedclass

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Question

Let $I = \int \sqrt{x^{2}+4 x+6} \, dx$ $= \int \sqrt{x^{2}+4 x+4+2} \, dx$ $= \int \sqrt{(x+2)^{2} + (\sqrt{2})^{2}} \, dx$ Using the standard integr

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Let $I = \int \sqrt{x^{2}+4 x+6} \, dx$
$= \int \sqrt{x^{2}+4 x+4+2} \, dx$
$= \int \sqrt{(x+2)^{2} + (\sqrt{2})^{2}} \, dx$
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 $(x+2)$ and $a = \sqrt{2}$:
$I = \frac{(x+2)}{2} \sqrt{(x+2)^{2} + 2} + \frac{2}{2} \log |(x+2) + \sqrt{(x+2)^{2} + 2}| + C$
$= \frac{(x+2)}{2} \sqrt{x^{2}+4 x+6} + \log |(x+2) + \sqrt{x^{2}+4 x+6}| + C$,where $C$ is an arbitrary constant.

Integrate the function: $\int \sqrt{x^{2}+4 x+6} \, dx$

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