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December 20, 2014

December 20, 2014

Posted by **Ang** on Friday, February 23, 2007 at 9:05am.

The integral of “x squared” in the numerator and “x squared plus x minus 6” in the denominator.

S X2 / (X2 + x – 6) dx

Thanks!

That's a messy one. According to my table of integrals.

The answer is

-x/6 -(1/72)loge(x^2 +x -6)

+ S dx/(x^2 +x -6)

and the last term is another integral that involves an arctangent.

I am not going to attempt to derive it.

Thanks for trying.

I do have the solution, but I'm stuck on part of ariving at the destination.

x - 9/5 ln !x+3! + 4/5 ln !x-2! + C

The trick may require using the method of partial fractions, whereby

x^2/[(x+3)(x-2)] is rewritten as

1 + (6-x)/(x+3)(x-2)]

The "1" term will lead to the "x" term in the integral.

The technique of partial fractions is then applied to the (6-x)/(x+3)(x-2)] term, leaving you with two terms of the form

(ax + b)/(x + 3) + (cx + d)/(x-2)

Do a search online or in your textbook on the method of parital fractions for additional help.

Thank you! Yes it was the algebraic partial fractions. YES!

Thanks again.

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