How do you express $(4x^2+6x-2)/((x-1)(x+1)^2)$ in partial fractions?

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Answer 1 · 2018-07-06T20:26:31

$(4x^2+6x-2)/((x-1)(x+1)^2)=2/(x+1)+2/(x+1)^2+2/(x-1)$

We consider the following ansatz

$(4x^2+6x-2)/((x-1)*(x+1)^2)=A/(x-1)+B/(x+1)+C/(x+1)^2$
multiplying by

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

we get

$4x^2+6x-2=A(x+1)^2+B(x+1)(x-1)+C(x-1)$

ultiplying out and rearranging

we get

$4x^2+6x-2=x^2(A+B)+x(2A+C)+C(x-1)$
this leads us to the following System

$A+B=4$

$2A+C=6$

$A-B-C=-2$

solving this System we get $A=B=C=2$

How do you express (4x^2+6x-2)/((x-1)(x+1)^2)(4x^2+6x-2)/((x-1)(x+1)^2) in partial fractions?