# Indefinite Integral

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• November 19th 2007, 09:44 AM
Del
Indefinite Integral
https://webwork.math.lsu.edu/webwork...1f13830961.png https://webwork.math.lsu.edu/webwork...a2ab21bf01.png

I've tried this so many times and I can't seem to find the right answer, please help.
• November 19th 2007, 09:45 AM
Krizalid
All you need to set it is $u=\sqrt{36x^2-1}.$
• November 19th 2007, 10:04 AM
Del
What would du be in that case?
• November 19th 2007, 10:56 AM
Krizalid
$du=\frac{36x}{\sqrt{36x^2-1}}\,dx.$