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.

Printable View

- Nov 19th 2007, 09:44 AMDelIndefinite 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. - Nov 19th 2007, 09:45 AMKrizalid
All you need to set it is $\displaystyle u=\sqrt{36x^2-1}.$

- Nov 19th 2007, 10:04 AMDel
What would du be in that case?

- Nov 19th 2007, 10:56 AMKrizalid
$\displaystyle du=\frac{36x}{\sqrt{36x^2-1}}\,dx.$