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mathz

Name: Anonymous 2008-09-08 22:12

how to integrate \int_1^9\dfrac{\sqrt(u)-2u^2}{u} du?

Name: Anonymous 2008-09-08 22:48

fail

Name: Anonymous 2008-09-08 23:14

Name: Anonymous 2008-09-08 23:39

Presumably he meant \int_1^9{{\sqrt(u)-2u^2} \over u} du.

Name: Anonymous 2008-09-09 10:10

>>4
yes! how do you do it?

Name: 4tran 2008-09-09 21:07

= \int_1^9{ \frac{1}{\sqrt(u)} - 2u }du
right?
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