Question
$\int_0^{2/3} {\frac{{dx}}{{4 + 9{x^2}}} = } $
$ = \frac{1}{9} \times \frac{1}{{2/3}}\left( {{{\tan }^{ - 1}}\frac{x}{{2/3}}} \right)_0^{2/3} $
$= \frac{\pi }{4} \times \frac{1}{6} = \frac{\pi }{{24}}$.
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