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Here is the problem: $\int \frac{5x^2-2}{x^2-4x-12}dx$

I factored it and got this form:

$\int \frac{5x^2-2}{(x-6)(x+2)}dx$ however the solution shows the next step looking like this: $\int 5+ \frac{20x+58}{(x-6)(x+2)}dx$

I know how to integrate and solve it, but this particular step has me confused. Can you help me to explain how they got there?

The reason I'm confused is because the following steps show the actual decomposition using A, B, etc over their respective denominators.

Edit: Now that I look at it, it looks like they divided it, but I don't understand why there is still a decomposition if the problem can be divided.

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You must first do polynomial division because for a partial fraction decomposition, you need the degree of the numerator smaller than that of the denominator. – Daniel Fischer Oct 17 '13 at 21:30
@DanielFischer this is my first problem where polynomial division is a prerequisite for the partial fraction decomposition. Thanks for that tip! – free_mind Oct 17 '13 at 21:32

In light of Daniel's comment, lets work on the integrand. I am making the final parts stepwise, so you can check the way well. You have $$\frac{5x^2-2}{x^2-4x-12} \tag{1}$$ $$=\frac{5x^2-2+\color{red}{20x}-\color{blue}{20x}+\color{brown}{60}-\color{yellow}{60}}{x^2-4x-12} \tag{2}$$ $$=\frac{5(x^2-\color{blue}{4x}-\color{yellow}{12})+(\color{red}{20x}+58)}{x^2-4x-12}\tag{3}$$ $$=5+2\times\frac{10x+29}{x^2-4x-12}\tag{4}$$ Now you have to do the partial fraction method because the degree of the nominator is less than the degree of the denominator. Indeed, we have to find constants $A,B$ such that: $$\frac{10x+29}{x^2-4x-12}=\frac{A}{x-6}+\frac{B}{x+2}$$

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At last you're back to Mathematics Exchange! @Babak S. – Maths Lover Aug 3 '14 at 13:44
@MathsLover: Thanks for saying so. ;-) – Babak S. Aug 3 '14 at 15:28
I really mean it:) I was looking into your account from time to time to see if you came back or not. I have left the site from November to May and when I was back, I didn't find you. I had a fear that you aren't ok. But now,I'm happy to see you here again:) never go away again, I hope. – Maths Lover Aug 3 '14 at 15:38
@MathsLover: It's very kind of you. Honestly, i am not back completely here cause of the preparing for an difficult exam. You are one of my best friend (brother) Fawzy. :-) – Babak S. Aug 3 '14 at 15:42
I wish you best of luck with your exams, I hope to see you here at least from time to time ;). It's a pleasure to be your friend ^_^. – Maths Lover Aug 3 '14 at 16:07

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