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So I have this question that looks like

$$ \frac{x^3 + 3x^2 - x - 8}{x^2 + x - 6} $$

and first I got the partial fraction so getting

$$ x + 2 + \frac{3x + 4}{x^2 + x -6} $$

but now I'm trying to integrate it and I cannot remember for the life of me how I should integrate the fraction on the end. Please help.

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  • $\begingroup$ I may be wrong but should the leading $3x+2$ actually be $x+2$? $\endgroup$ – illysial Nov 18 '14 at 17:04
  • $\begingroup$ Yes seems that I need to be more careful when tying in my questions $\endgroup$ – Paul Nov 18 '14 at 17:24
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Hint:

Remember that $x^2+x-6=(x-2)(x+3)$.

Now apply the partial fraction decomposition again:

$$\frac{3x+4}{x^2+x-6}=\frac{3x+4}{(x-2)(x+3)}=\frac{A}{x-2}+\frac{B}{x+3}$$

Also, it seems as if your division is not correct:

$$\require{cancel}\frac{x^3 + 3x^2 - x - 8}{x^2 + x - 6}=\color{red}{\cancel{3}}x+2+\frac{3x+4}{x^2+x-6}$$

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  • $\begingroup$ Would you believe that I actually have that factorisation on my page here, I just didn't see that I should do the partial fractions again. Don't you just feel stupid sometimes :) $\endgroup$ – Paul Nov 18 '14 at 17:01
  • $\begingroup$ Yeah seems that's a mistype from an earlier mistake that I have scored out. I have got the correct answer now $\endgroup$ – Paul Nov 18 '14 at 17:23
  • $\begingroup$ Alright, don't forget to press the accept button! ;-) $\endgroup$ – rae306 Nov 18 '14 at 17:52
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$$3x+2+\frac{3x+4}{x^2+x−6} =\\3x+2+\frac{3x+4}{(x-2)(x+3)} =\\ 3x+2+\frac{a}{(x-2)}+\frac{b}{(x-2)} =\\ $$now find a,b $$\frac{a}{(x-2)}+\frac{b}{(x-2)} =\frac{a(x+3)+b(x-2)}{(x-2)(x+3)}=\frac{3x +4}{(x-2)(x+3)}\\\rightarrow \\(a+b)x=3x\\3a-2b=4\\a=2,b=1\\ $$

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You can factor the denominator of the last term and decompose again:

$$\frac{3x+4}{x^2+x-6} = \frac{3x+4}{(x+3)(x-2)} = \frac{1}{x+3} + \frac{2}{x-2}.$$

Can you take it from there?

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