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Calcolatore di decomposizione in fratti semplici

Trova la scomposizione in fratti semplici passo dopo passo

Questo calcolatore online calcolerà la decomposizione in fratti semplici della funzione razionale, con i passaggi mostrati.

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Solution

Your input: perform the partial fraction decomposition of $$$\frac{1}{x^{4} - 9}$$$

Factor the denominator: $$$\frac{1}{x^{4} - 9}=\frac{1}{\left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right) \left(x^{2} + 3\right)}$$$

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

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

Expand the right-hand side:

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

Collect up the like terms:

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

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

$$\begin{cases} A + C + D = 0\\B - \sqrt{3} C + \sqrt{3} D = 0\\- 3 A + 3 C + 3 D = 0\\- 3 B - 3 \sqrt{3} C + 3 \sqrt{3} D = 1 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=0$$$, $$$B=- \frac{1}{6}$$$, $$$C=- \frac{\sqrt{3}}{36}$$$, $$$D=\frac{\sqrt{3}}{36}$$$

Therefore,

$$\frac{1}{\left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right) \left(x^{2} + 3\right)}=\frac{- \frac{1}{6}}{x^{2} + 3}+\frac{- \frac{\sqrt{3}}{36}}{x + \sqrt{3}}+\frac{\frac{\sqrt{3}}{36}}{x - \sqrt{3}}$$

Answer: $$$\frac{1}{x^{4} - 9}=\frac{- \frac{1}{6}}{x^{2} + 3}+\frac{- \frac{\sqrt{3}}{36}}{x + \sqrt{3}}+\frac{\frac{\sqrt{3}}{36}}{x - \sqrt{3}}$$$