How do you express #(4x-2) /( 3(x-1)^2)# in partial fractions?

Question

2808 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-23T07:29:16

The answer is #=(2/3)/(x-1)^2+(4/3)/(x-1)#

Let's do the decomposition into partial fractions

#(4x-2)/(3(x-1)^2)=1/3(A/(x-1)^2+B/(x-1))#

#=1/3(A+B(x-1))/(x-1)^2#

Therefore,

#4x-2=A+B(x-1)#

Comparing the coefficients of x,

#4=B#

Let #x=1#

#2=A#

Therefore,

#(4x-2)/(3(x-1)^2)=1/3(2/(x-1)^2+4/(x-1))#