# Perfect Squares-Partial Fractions

Regarding partial fractions, where rational functions, whose denominator can be factorised into degree 2 polynomials with complex roots such as $\frac{2}{(x+1)(x^2+4)}$, can you evaluate the numerators in the decomposed fraction for the quadratic factor as follows?

$$\frac{2}{(x+1)(x^2+4)} = \frac{A}{x+1}+\frac{B}{x+2i}+\frac{C}{x-2i}$$

$$A=\lim_{x \to -1} \frac{2}{x^2+4}$$ $$B=\lim_{x \to 2i} \frac{2}{(x+1)(x-2i)}$$ $$C=\lim_{x \to -2i} \frac{2}{(x+1)(x+2i)}$$

• You can, but its better to limit to $$\dfrac A{x+1}+\dfrac{Bx+C}{x^2+4}$$ – lab bhattacharjee Jun 5 '17 at 7:23

Starting with $$\frac{2}{(x+1)(x^2+4)} = \frac{A}{x+1}+\frac{B}{x+2i}+\frac{C}{x-2i}$$ just reduce to same denominator to write $$2=A(x+2i)(x-2i)+B(x+1)(x-2i)+C(x+1)(x+2i)$$ Since this is valid for all $x$, just set successively, as you did, $x=-1$, $x=2i$, $x=-2i$. To me, this looks simpler than using limits but the general underlying idea is good.