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

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Answer 1 · 2016-02-07T13:49:41

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

Let $x/((x-1)(x+1))=(A)/(x-1)+(B)/(x+1)$

$x/((x-1)(x+1))=(A)/(x-1)((x+1)/(x+1))+(B)/(x+1)*((x-1)/(x-1))$

$x/((x-1)(x+1))=(Ax+A+Bx-B)/((x-1)(x+1))$

Equate the numerators

$x=Ax+A+Bx-B$

$x=Ax+Bx+A-B$

$1*x+0=(A+B)x+(A-B)$

so that

$A+B=1$ and $A-B=0$

solving for A and B

$A=1/2$ and $B=1/2$

so therefore

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

have a nice day ! from the Philippines..

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