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December 20, 2014

December 20, 2014

Posted by **Frank** on Friday, October 15, 2010 at 2:04am.

- calculus -
**drwls**, Friday, October 15, 2010 at 7:42amLet 2x^3+15x^2+6x+23 = u

du = 6x^2 + 30x + 6

= 6(x^2 + 5x + 1)dx

You got lucky with this one!

Therefore the integrand can be written

(1/6)du/u^1/2

and the integral is

(1/3)*u^1/2

= (1/3)(2x^3+15x^2+6x+23)^1/2

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