# Indefinite Integral

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

All you need to set it is $u=\sqrt{36x^2-1}.$
$du=\frac{36x}{\sqrt{36x^2-1}}\,dx.$