Calculadora de descomposición en fracciones parciales

Your input: perform the partial fraction decomposition of $$$\frac{1}{x^{2} + x - 2}$$$

Factor the denominator: $$$\frac{1}{x^{2} + x - 2}=\frac{1}{\left(x - 1\right) \left(x + 2\right)}$$$

The form of the partial fraction decomposition is

$$\frac{1}{\left(x - 1\right) \left(x + 2\right)}=\frac{A}{x - 1}+\frac{B}{x + 2}$$

Write the right-hand side as a single fraction:

$$\frac{1}{\left(x - 1\right) \left(x + 2\right)}=\frac{\left(x - 1\right) B + \left(x + 2\right) A}{\left(x - 1\right) \left(x + 2\right)}$$

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

$$1=\left(x - 1\right) B + \left(x + 2\right) A$$

Expand the right-hand side:

$$1=x A + x B + 2 A - B$$

Collect up the like terms:

$$1=x \left(A + B\right) + 2 A - B$$

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

$$\begin{cases} A + B = 0\\2 A - B = 1 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=\frac{1}{3}$$$, $$$B=- \frac{1}{3}$$$

Therefore,

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

Answer: $$$\frac{1}{x^{2} + x - 2}=\frac{\frac{1}{3}}{x - 1}+\frac{- \frac{1}{3}}{x + 2}$$$