Question
Evaluate the following integrals:
$\int\frac{1}{3\sqrt{\text{x}^2}}\text{dx}$

Answer

$\int\frac{\text{dx}}{3\sqrt{\text{x}^2}}$
$=\int\frac{\text{dx}}{\text{x}^\frac{2}{3}}$
$=\int\text{x}^\frac{-2}{3}\text{dx}$
$=\frac{\text{x}^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}+\text{c}$
$=3\text{x}^\frac{1}{3}+\text{c}$

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