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Calculatrice de décomposition en fractions partielles
Trouver la décomposition en fractions partielles étape par étape
Cette calculatrice en ligne trouvera la décomposition en éléments simples de la fonction rationnelle, en montrant les étapes.
Solution
Your input: perform the partial fraction decomposition of $$$\frac{2 x^{2} - 1}{x^{6} + 1}$$$
Factor the denominator: $$$\frac{2 x^{2} - 1}{x^{6} + 1}=\frac{2 x^{2} - 1}{\left(x^{2} + 1\right) \left(x^{2} - \sqrt{3} x + 1\right) \left(x^{2} + \sqrt{3} x + 1\right)}$$$
The form of the partial fraction decomposition is
$$\frac{2 x^{2} - 1}{\left(x^{2} + 1\right) \left(x^{2} - \sqrt{3} x + 1\right) \left(x^{2} + \sqrt{3} x + 1\right)}=\frac{A x + B}{x^{2} + 1}+\frac{C x + D}{x^{2} + \sqrt{3} x + 1}+\frac{E x + F}{x^{2} - \sqrt{3} x + 1}$$
Write the right-hand side as a single fraction:
$$\frac{2 x^{2} - 1}{\left(x^{2} + 1\right) \left(x^{2} - \sqrt{3} x + 1\right) \left(x^{2} + \sqrt{3} x + 1\right)}=\frac{\left(x^{2} + 1\right) \left(x^{2} - \sqrt{3} x + 1\right) \left(C x + D\right) + \left(x^{2} + 1\right) \left(x^{2} + \sqrt{3} x + 1\right) \left(E x + F\right) + \left(x^{2} - \sqrt{3} x + 1\right) \left(x^{2} + \sqrt{3} x + 1\right) \left(A x + B\right)}{\left(x^{2} + 1\right) \left(x^{2} - \sqrt{3} x + 1\right) \left(x^{2} + \sqrt{3} x + 1\right)}$$
The denominators are equal, so we require the equality of the numerators:
$$2 x^{2} - 1=\left(x^{2} + 1\right) \left(x^{2} - \sqrt{3} x + 1\right) \left(C x + D\right) + \left(x^{2} + 1\right) \left(x^{2} + \sqrt{3} x + 1\right) \left(E x + F\right) + \left(x^{2} - \sqrt{3} x + 1\right) \left(x^{2} + \sqrt{3} x + 1\right) \left(A x + B\right)$$
Expand the right-hand side:
$$2 x^{2} - 1=x^{5} A + x^{5} C + x^{5} E + x^{4} B - \sqrt{3} x^{4} C + x^{4} D + \sqrt{3} x^{4} E + x^{4} F - x^{3} A + 2 x^{3} C - \sqrt{3} x^{3} D + 2 x^{3} E + \sqrt{3} x^{3} F - x^{2} B - \sqrt{3} x^{2} C + 2 x^{2} D + \sqrt{3} x^{2} E + 2 x^{2} F + x A + x C - \sqrt{3} x D + x E + \sqrt{3} x F + B + D + F$$
Collect up the like terms:
$$2 x^{2} - 1=x^{5} \left(A + C + E\right) + x^{4} \left(B - \sqrt{3} C + D + \sqrt{3} E + F\right) + x^{3} \left(- A + 2 C - \sqrt{3} D + 2 E + \sqrt{3} F\right) + x^{2} \left(- B - \sqrt{3} C + 2 D + \sqrt{3} E + 2 F\right) + x \left(A + C - \sqrt{3} D + E + \sqrt{3} F\right) + B + D + F$$
The coefficients near the like terms should be equal, so the following system is obtained:
$$\begin{cases} A + C + E = 0\\B - \sqrt{3} C + D + \sqrt{3} E + F = 0\\- A + 2 C - \sqrt{3} D + 2 E + \sqrt{3} F = 0\\- B - \sqrt{3} C + 2 D + \sqrt{3} E + 2 F = 2\\A + C - \sqrt{3} D + E + \sqrt{3} F = 0\\B + D + F = -1 \end{cases}$$
Solving it (for steps, see system of equations calculator), we get that $$$A=0$$$, $$$B=-1$$$, $$$C=- \frac{\sqrt{3}}{6}$$$, $$$D=0$$$, $$$E=\frac{\sqrt{3}}{6}$$$, $$$F=0$$$
Therefore,
$$\frac{2 x^{2} - 1}{\left(x^{2} + 1\right) \left(x^{2} - \sqrt{3} x + 1\right) \left(x^{2} + \sqrt{3} x + 1\right)}=\frac{-1}{x^{2} + 1}+\frac{- \frac{\sqrt{3} x}{6}}{x^{2} + \sqrt{3} x + 1}+\frac{\frac{\sqrt{3} x}{6}}{x^{2} - \sqrt{3} x + 1}$$
Answer: $$$\frac{2 x^{2} - 1}{x^{6} + 1}=\frac{-1}{x^{2} + 1}+\frac{- \frac{\sqrt{3} x}{6}}{x^{2} + \sqrt{3} x + 1}+\frac{\frac{\sqrt{3} x}{6}}{x^{2} - \sqrt{3} x + 1}$$$