How do you express $\frac{1}{x^{3} - 4 x}$ $1/(x^3-4x)$ in partial fractions?

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How do you express $\frac{1}{x^{3} - 4 x}$ $1/(x^3-4x)$ in partial fractions?

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Answer 1 · 2016-02-11T00:03:04

You can analyze it as follows

$\frac{1}{x^{3} - 4 x} = \frac{1}{x (x - 2) (x + 2)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2}$ $1/[x^3-4x]=1/(x(x-2)(x+2))=A/x+B/(x-2)+C/(x+2)$

Now try various values of $x$ $x$ except $0, - 2, 2$ $0,-2,2$ to find the values of
constants $A, B, C$ $A,B,C$ finally you get

$\frac{1}{x^{3} - 4 x} = - \frac{1}{4 x} + \frac{1}{8 \cdot (x - 2)} + \frac{1}{8 \cdot (x + 2)}$ $1/[x^3-4x]=-1/(4x)+1/[8*(x-2)]+1/[8*(x+2)]$

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How do you express $\frac{1}{x^{3} - 4 x}$ $1/(x^3-4x)$ in partial fractions?

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