# Partial Fraction and the Integration

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.

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

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:

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

• 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 :) – Paul Nov 18 '14 at 17:01
• Yeah seems that's a mistype from an earlier mistake that I have scored out. I have got the correct answer now – Paul Nov 18 '14 at 17:23
• Alright, don't forget to press the accept button! ;-) – rae306 Nov 18 '14 at 17:52

$$3x+2+\frac{3x+4}{x^2+x−6} =\\3x+2+\frac{3x+4}{(x-2)(x+3)} =\\ 3x+2+\frac{a}{(x-2)}+\frac{b}{(x-2)} =\\$$now find a,b $$\frac{a}{(x-2)}+\frac{b}{(x-2)} =\frac{a(x+3)+b(x-2)}{(x-2)(x+3)}=\frac{3x +4}{(x-2)(x+3)}\\\rightarrow \\(a+b)x=3x\\3a-2b=4\\a=2,b=1\\$$

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?