Friday

December 19, 2014

December 19, 2014

Posted by **Annie** on Sunday, May 10, 2009 at 9:06pm.

- calculus -
**drwls**, Sunday, May 10, 2009 at 11:52pmdy = (3x)/(1-x^2) dx

y = Integral of [3x/(1-x^2)] dx + constant

For the integration, let

1-x^2 = u

-2x dx = du

Which make the integral

Integral of (-3/2)du/u = -(3/2)ln(1-x^2)

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