How do you use partial fraction decomposition to decompose the fraction to integrate #(4x-2)/[3(x-1)^2]#?

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Answer 1 · 2015-06-15T21:41:51

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

#(4x-2)/[3(x-1)^2] = 1/3 * (4x-2)/(x-1)^2#

So we really only need to decompose the second factor:

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

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

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

So we need:

#Ax-A+B = 4x-2#.

So #A=4# and, since #-A+B = -2#, we get #B=2#

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

#int (4x-2)/[3(x-1)^2] dx = 1/3 [int 4/(x-1) dx+ int2/(x-1)^2 dx]#