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Question
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Solution
\[\int_0^3 \frac{1}{x^2 + 9} d x\]
\[ = \int_0^3 \frac{1}{x^2 + 3^2} d x\]
\[ = \frac{1}{3} \left[ \tan^{- 1} \frac{x}{3} \right]_0^3 \]
\[ = \frac{1}{3}\left( \tan^{- 1} 1 - \tan^{- 1} 0 \right)\]
\[ = \frac{1}{3}\left( \frac{\pi}{4} - 0 \right)\]
\[ = \frac{\pi}{12}\]
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