MHB Why Use Ax + B in Partial Fraction Decomposition?

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In partial fraction decomposition, when dealing with an irreducible quadratic factor in the denominator, the numerator should be a polynomial of one degree less, typically represented as Ax + B. This is essential for correctly expressing the integral, as seen in the example provided. The reasoning behind this is that a linear polynomial captures all possible behaviors of the function over the quadratic factor. Remembering this rule can help avoid common mistakes, such as incorrectly using Ax + Bx or (A + B)/denominator. Understanding this concept is crucial for accurate integration in calculus.
shamieh
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When I'm evaluating a problem like
$$
\int \frac{2x^2 + 8x + 9}{(x^2 + 2x + 5)(x + 2)} = \frac{Ax + B}{x^2 + 2x + 5} + \frac{C}{x+2}$$

I understand how to get the C part, that's simple. But what is a Good trick to know that I need to have $$Ax + B$$ over the $$x^2 + 2x + 5$$ denominator? Is there a way I can remember it easier? Because sometimes I will mistakenly put $$Ax + Bx$$ or $$(A + B)$$/denominator + $$C$$/denominator

Thanks for your time
 
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If you have an irreducible factor in the denominator, then in the partial fraction decomposition, you want to put a polynomial one degree less than this factor in the numerator.

So, if for example, you have an irreducible quadratic factor in the denominator, you want to put a general linear polynomial over this factor in the decomposition. The general linear polynomial is of the form $Ax+B$.
 
Awesome, thanks Mark.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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