The number partial fractions of $\frac{1}{x^3\left(x+2\right)}$  is

$\Rightarrow 1=A{x}^2\left(x+2\right)+Bx\left(x+2\right)+C\left(x+2\right)+D{x}^3$

$A+D=0;2A+B=0;2B+C=0;2C=1\Rightarrow C=\frac{1}{2}$

$2B+C=0\Rightarrow B=-\frac{1}{4}$

$2A+B=0\Rightarrow A=\frac{1}{8}$

$A+D=0\Rightarrow D=-\frac{1}{8}$

$\therefore \frac{1}{x^3\left(x+2\right)}=\frac{1}{8x}-\frac{1}{4x^2}+\frac{1}{2x^3}-\frac{1}{8\left(x+2\right)}$

The number partial fractions of $\frac{1}{x^3\left(x+2\right)}$  is 4