Question
Find $\int \frac{\text{d}x}{x^{2} + 4x + 8}$

Answer

$\int \frac{\text{dx}}{\text{x}^{2} + \text{4x + 8}} = \int \frac{\text{dx}}{(\text {x + 2)}^{2} + (2)^{2}}$
$= \frac{1}{2} \tan^{-1} \frac{\text{x + 2}}{2} + \text{C}$

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