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goformit100 Group TitleBest ResponseYou've already chosen the best response.0
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 11 months ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Try substituting the bottom
 11 months ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.0
\[\int\frac{\sec^23x}{1+4\tan3x}~dx\] Let \(u=1+4\tan 3x\), so \(\dfrac{1}{12}du=\sec^23x~dx\): \[\frac{1}{12}\int\frac{du}{u}\]
 11 months ago
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