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

Question

1706 views around the world

You can reuse this answer
Creative Commons License

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#