Question

Resolve $\frac{x}{1+x^3}$ into partial fractions.

Answer 1

Factor of $1+x^3=(1+x)(1-x+x^2)$
$\therefore$ $\frac{x}{1+x^3}=\frac{x}{(1+x)(1-x+x^2)}$
Let $\frac{x}{(1+x)(1-x+x^2)}=\frac{A}{1+x}+\frac{Bx+C}{1-x+x^2}....\left(A\right)$
$x=A(1-x+x^2)+(Bx+C)(1+x)$
$x=A-Ax+A{x}^2+Bx+B{x}^2+C+Cx$
Comparing the coefficients of $x,x^2$ and constant terms both sides
$A+B=\mathrm{0......}\left(1\right)$
$-A+B+C=\mathrm{1........}\left(2\right)$
$A+C=\mathrm{0........}\left(3\right)$
equation (1)-(3)
$A+B=\mathrm{0.......}\left(1\right)$
$A+C=\mathrm{0......}\left(3\right)$
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$B-C=\mathrm{0.......}\left(4\right)$
equation (2)-(4)
$-A+B+C=1$
$B-C=0$
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$-A+2B=\mathrm{1.......}\left(5\right)$
equation (1)+equation (5)
$A+B=\mathrm{0......}\left(1\right)$
$-A+2B=\mathrm{1......}\left(5\right)$
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$3B=1$
$B=\frac{1}{3}$
by equation (1) $A+B=0\Rightarrow A=-\frac{1}{3}$
by equation (3) $A+C=0\Rightarrow C=\frac{1}{3}$
$\frac{x}{1+x^3}=\frac{-1}{3(1-x)}+\frac{\frac{1}{3}x+\frac{1}{3}}{1-x+x^2}$
$=\frac{1}{3(1-x)}+\frac{x+1}{3(1-x+x^2)}$