Calculadora de Decomposição em Frações Parciais

Your input: perform the partial fraction decomposition of $$$\frac{1}{x^{3} \left(x + 1\right)}$$$

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

The denominators are equal, so we require the equality of the numerators:

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

Expand the right-hand side:

$$1=x^{3} A + x^{3} D + x^{2} A + x^{2} B + x B + x C + C$$

Collect up the like terms:

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

The coefficients near the like terms should be equal, so the following system is obtained:

$$\begin{cases} A + D = 0\\A + B = 0\\B + C = 0\\C = 1 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=1$$$, $$$B=-1$$$, $$$C=1$$$, $$$D=-1$$$

Therefore,

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

Answer: $$$\frac{1}{x^{3} \left(x + 1\right)}=\frac{1}{x}+\frac{-1}{x^{2}}+\frac{1}{x^{3}}+\frac{-1}{x + 1}$$$