Here's the question you clicked on:
Zaira08
integrate integral sec^2 3 xdx /1+4tan3x
A warm Welcome to OpenStudy. I can help guide you through this useful site. You can ask your questions to me or you can message me. Please use the chat for off topic questions. Remember to give the person who helped you a medal by clicking on "Best Answer". We follow a code of conduct, ( http://openstudy.com/code-of-conduct ). Please take a moment to read it.
Try substituting the bottom
\[\int\frac{\sec^23x}{1+4\tan3x}~dx\] Let \(u=1+4\tan 3x\), so \(\dfrac{1}{12}du=\sec^23x~dx\): \[\frac{1}{12}\int\frac{du}{u}\]