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Question
Evaluate: \[\int\frac{1}{x^2 + 16}\text{ dx }\]
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Solution
\[ = \frac{1}{4} \tan^{- 1} \frac{x}{4} + c\]
\[\text{ Hence,} \int\frac{1}{x^2 + 16}dx = \frac{1}{4} \tan^{- 1} \frac{x}{4} + c\]
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