Why do we need a $Bx$ term for partial fractions with irreducible quadratic factors?

Question

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

Answer 1

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 $$ \frac{\beta}{x+i}+\frac{\gamma}{x-i} = \frac{\beta(x-i)+\gamma(x+i)}{x^2+1} = \frac{B x + C}{x^2+1} $$

Answer 2

$$f(x)=\frac{1}{x(x^2+1)}=x\cdot\frac{1}{x^2(x^2+1)}=x\left(\frac{1}{x^2}-\frac{1}{x^2+1}\right)=\frac{1}{x}-\frac{x}{x^2+1}.$$

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 $f(x)-\frac{1}{x}$.