How do you express $\frac{x^{2} - 1}{x (x^{2} + 1)}$ $(x^(2) - 1) / (x(x^(2)+1))$ in partial fractions?

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Answer 1 · 2016-11-26T11:18:52

The answer is $= - \frac{1}{x} + \frac{2 x}{x^{2} + 1}$ $=-1/(x)+(2x)/(x^2+1)$

Let's do the decomposition into partial fractions

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

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

Therefore,

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

Let $x = 0$ $x=0$ , $\Rightarrow$ $=>$ $- 1 = A$ $-1=A$

Coefficients of $x^{2}$ $x^2$

$1 = A + B$ $1=A+B$ , $\Rightarrow$ $=>$ , $B = 1 - A = 2$ $B=1-A=2$

coefficients of $x$ $x$

$0 = C$ $0=C$

Finally, we have

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

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