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

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Answer 1 · 2016-08-14T00:06:50

$\frac{x - 1}{x^{3} + x} = \frac{x + 1}{x^{2} + 1} - \frac{1}{x}$ $(x-1)/(x^3+x) = (x+1)/(x^2+1) - 1/x$

$\frac{x - 1}{x^{3} + x} = \frac{x - 1}{x (x^{2} + 1)}$ $(x-1)/(x^3+x) = (x-1)/(x(x^2+1))$

$\frac{x - 1}{x (x^{2} + 1)} = \frac{A}{x} + \frac{B x + C}{x^{2} + 1}$ $(x-1)/(x(x^2+1)) = A/x + (Bx+C)/(x^2+1)$

$x - 1 = A (x^{2} + 1) + x (B x + C)$ $x-1 = A(x^2+1) + x(Bx+C)$

$x - 1 = A x^{2} + A + B x^{2} + C x$ $x-1 = Ax^2 + A + Bx^2 + Cx$

Comparing coefficients:

$0 = A + B$ $0 = A + B$

$1 = C$ $1 = C$

$- 1 = A \Rightarrow B = 1$ $-1 = A implies B = 1$

$\frac{x - 1}{x^{3} + x} = \frac{x + 1}{x^{2} + 1} - \frac{1}{x}$ $(x-1)/(x^3+x) = (x+1)/(x^2+1) - 1/x$

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