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Kalkulator Penguraian Pecahan Parsial

Temukan dekomposisi pecahan parsial langkah demi langkah

Kalkulator daring ini akan menentukan dekomposisi pecahan parsial dari fungsi rasional, beserta langkah-langkahnya.

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Solution

Your input: perform the partial fraction decomposition of $$$\frac{1}{u^{3} \left(1 - u^{2}\right)}$$$

Simplify the expression: $$$\frac{1}{u^{3} \left(1 - u^{2}\right)}=\frac{-1}{u^{5} - u^{3}}$$$

Factor the denominator: $$$\frac{-1}{u^{5} - u^{3}}=\frac{-1}{u^{3} \left(u - 1\right) \left(u + 1\right)}$$$

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

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

Expand the right-hand side:

$$-1=u^{4} A + u^{4} D + u^{4} E + u^{3} B - u^{3} D + u^{3} E - u^{2} A + u^{2} C - u B - C$$

Collect up the like terms:

$$-1=u^{4} \left(A + D + E\right) + u^{3} \left(B - D + E\right) + u^{2} \left(- A + C\right) - u B - C$$

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

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

Solving it (for steps, see system of equations calculator), we get that $$$A=1$$$, $$$B=0$$$, $$$C=1$$$, $$$D=- \frac{1}{2}$$$, $$$E=- \frac{1}{2}$$$

Therefore,

$$\frac{-1}{u^{3} \left(u - 1\right) \left(u + 1\right)}=\frac{1}{u}+\frac{0}{u^{2}}+\frac{1}{u^{3}}+\frac{- \frac{1}{2}}{u + 1}+\frac{- \frac{1}{2}}{u - 1}=\frac{1}{u}+\frac{1}{u^{3}}+\frac{- \frac{1}{2}}{u + 1}+\frac{- \frac{1}{2}}{u - 1}$$

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


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