Posted by Ang on .
Having trouble getting the correct solution.
The integral of “x squared” in the numerator and “x squared plus x minus 6” in the denominator.
S X2 / (X2 + x – 6) dx
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!