Alg5.1.11

Write $\frac{1}{x^3(x+2)}\,$ as the sum of partial fractions.

Let $\frac{1}{x^3(x+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+2}\,$

$1=A(x^2)(x+2)+B(x)(x+2)+C(x+2)+D x^3\,$

Substituting $x=0\,$ in the above,we get

$1=2C,C=\frac{1}{2}\,$

Substituting $x=-2\,$

$1=D(-2)^3,D=-\frac{1}{8}\,$

Comparing the coefficients of $x^3\,$ on both sides,we get $0=A+D,A=\frac{1}{8}\,$

Comparing the coefficients of $x^2\,$ on both sides,we get $0=2A+B,B=-\frac{1}{4}\,$

Therefore, $\frac{1}{x^3(x+2)}=\frac{1}{8x}-\frac{1}{4x^2}+\frac{1}{2x^3}-\frac{1}{8(x+2)}\,$