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

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

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Answer 1 · 2016-02-08T19:19:43

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

Explanation:

We should solve the equation:

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

Putting everithing with the samw denominators:

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

Then we can eliminate denominators:

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

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

The we have three equations to solve:

Terms in $x^{2} : 1 = A + B$ $x^2: 1=A+B$

Terms in $x : 0 = C$ $x: 0=C$

Terms in $1 : - 1 = A$ $1: -1=A$

The solution is: A=-1, B=2, C=0.

So,

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

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

1 Answer

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