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Why is there Bx+c term when we try to split partial fraction with irreducible quadratic?

Eg:

$$\frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$$

I think that splitting partial fraction is intuition when we directly put it as $\frac{A}{linear factor}$ but I don't understand why we have to introduce a $Bx$ term for quadratic factors

tryst with freedom
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2 Answers2

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It's instructive to see what happens if you omit it. Suppose we had $$ \frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{B}{x^2+1} $$Cross multiplying, we get $$ 1 = A (x^2+1) + B(x) $$Matching the constant terms, we see $A=1$; matching the quadratic terms, we see $A=0$. This is a contradiction, so our decomposition is invalid. The need for a linear term is to balance the degree of the quadratic in the denominator.

If you were working over $\mathbb{C}$ instead of $\mathbb{R}$, you could split $x^2+1$ completely: $$ \frac{1}{x(x^2+1)} = \frac{1}{x(x+i)(x-i)} = \frac{A}{x}+\frac{\beta}{x+i}+\frac{\gamma}{x-i} $$Combining the last two terms gives $$ \frac{\beta}{x+i}+\frac{\gamma}{x-i} = \frac{\beta(x-i)+\gamma(x+i)}{x^2+1} = \frac{B x + C}{x^2+1} $$

Integrand
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  • You should get on the LHS $x$ and not $1$, when you cross multiply – user289143 Apr 29 '20 at 15:12
  • Umm so how would one know when to assume of the form Ax+B ? – Schwarz Kugelblitz Apr 29 '20 at 15:13
  • @SchwarzKugelblitz working over $\mathbb{R}$, whenever you have an irreducible quadratic. This happens when the discriminant [in $ax^2+bx+c$, the quantity $b^2-4ac$] is less than $0$. – Integrand Apr 29 '20 at 15:16
  • Would we assume Ax+B if the original denominator was of the form (dx^2+cx+f)^2? – Schwarz Kugelblitz Apr 29 '20 at 15:17
  • No; consider $(x^2-1)^2$, which we could factor as $(x-1)^2(x+1)^2$. The issue here is that you need multiple factors to account for the multiplicity; being irreducible over $\mathbb{R}$ is a different condition. – Integrand Apr 29 '20 at 15:19
  • I don't consider this answer satisfactory. It just goes and says that if you don't do this then you get the wrong answer... which doesn't shed insight onto the 'why?' of things. – tryst with freedom Apr 29 '20 at 15:25
  • That's a fair critique. Do you buy the second bit about splitting completely over $\mathbb{C}$? – Integrand Apr 29 '20 at 15:26
  • that's a good idea actually I didn't understand it when I first read it. I think both answers posted now have it's own insight in themselves. I'll wait and see if there is some other idea also left – tryst with freedom Apr 30 '20 at 00:53
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$$f(x)=\frac{1}{x(x^2+1)}=x\cdot\frac{1}{x^2(x^2+1)}=x\left(\frac{1}{x^2}-\frac{1}{x^2+1}\right)=\frac{1}{x}-\frac{x}{x^2+1}.$$

Also $$ A = \lim_{x\to 0} x f(x) = \lim_{x\to 0}\frac{1}{x^2+1} = 1$$ so $Bx+C$ can be found by simplifying $f(x)-\frac{1}{x}$.

Jack D'Aurizio
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