How do you express $(x^2-1)/((x)*(x^2+1) )$ in partial fractions?

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Answer 1 · 2016-02-08T19:19:43

$-1/x+(2x)/(x^2+1)$

We should solve the equation:

$(x^2-1)/(x(x^2+1))=A/x+(Bx+C)/(x^2+1)$

Putting everithing with the samw denominators:

$(x^2-1)/(x(x^2+1))=(A(x^2+1))/(x(x^2+1))+((Bx+C)x)/(x(x^2+1))$

Then we can eliminate denominators:

$x^2-1=(Ax^2+A)+(Bx^2+Cx)$

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

The we have three equations to solve:

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

Terms in $x: 0=C$

Terms in $1: -1=A$

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

So,

$(x^2-1)/(x(x^2+1))=-1/x+(2x)/(x^2+1)$

How do you express (x^2-1)/((x)*(x^2+1) )(x^2-1)/((x)*(x^2+1) ) in partial fractions?