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I understand that if we have a quadratic factor such as in

$\frac{8}{(x^2 + 1)(2x-3)} $

and we want to decompose, we should have a linear factor above $ x^2 +1$. Is the reason behind this primarily for calculus purposes, since if we have the numerator in linear form it will be easier to take the anti derivative? Theoretically, we could solve as if there is a constant in the numerator, though then we would have integration problems down the road?

user342258
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1 Answers1

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Hint. Observe that, we have $$ \frac{a}{x^2+1}+\frac{c}{2x-3}=\frac{c x^2+2ax+c-3 a}{\left(1+x^2\right)(2 x-3) } $$ and the latter fraction can not be equal to $$ \frac{8}{\left(1+x^2\right)(2 x-3) } $$ for all appropriate $x$. Do you see why?

Whereas, $$ \frac{8}{\left(1+x^2\right)(2 x-3) }=\frac{ax+b}{x^2+1}+\frac{c}{2x-3}=\frac{(2a+c) x^2+(2b-3a)x+c-3 b}{\left(1+x^2\right)(2 x-3) } $$ admits a valid solution: $$ a=-\frac{16}{13}, \quad b=-\frac{24}{13},\quad c=\frac{32}{13}. $$

Remark. Another way to see what happens, is to decompose in $\mathbb{C}(x)$: $$ \frac{8}{\left(1+x^2\right)(2 x-3) }=\frac{A}{x-i}+\frac{B}{x+i}+\frac{C}{2x-3}. $$

Olivier Oloa
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  • Thanks for the answer. Why can $ cx^2 + 2ax + c - 3x $ never equal 8? – Haim May 31 '16 at 00:07
  • If you want $c x^2+2ax+c-3 a$ to be constant for all $x$, you need that the coefficient of $x^2$ to be $0$ and the coefficient of $x$ to $0$, giving $c=0$ and $2a=0$ thus $c x^2+2ax+c-3 a$ reduces to $0$, it can't be equal to $8$ for all $x$. – Olivier Oloa May 31 '16 at 00:14