How do you integrate #(6x^2+1)/(x^2(x-1)^2)# using partial fractions?

1 Answer
Oct 22, 2016

The integral is #int((6x^2+1)dx)/(x^2(x-1)^2)=-1/x+2lnx-2ln(x-1)-7/(x-1) +C#

Explanation:

Let's first determine the coefficients of the partial fractions

#(6x^2+1)/(x^2(x-1)^2)=A/x^2+B/x+C/(x-1)+D/(x-1)^2#

#(6x^2+1)/(x^2(x-1)^2)=(A(x-1)^2+Bx(x-1)^2+Cx^2(x-1)+Dx^2)/(x^2(x-1)^2)#

#(6x^2+1)=A(x-1)^2+Bx(x-1)^2+Cx^2(x-1)+Dx^2#

Let x=0, then #1=A+0+0+0# ; So #A=1#
Let x=1, then #7=D# ; so #D=7#
Also,coefficients of #x^3#; #0=B+C#, ; so #B=-C#
And coefficients of #x^2# ; #6=A-2B-C+D#
So #6=1+2C-C+7# #=>##C=-2#
And #B=2#

#(6x^2+1)/(x^2(x-1)^2)=1/x^2+2/x-2/(x-1)+7/(x-1)^2#

So

#int((6x^2+1)dx)/(x^2(x-1)^2)=int(1dx)/x^2+int(2dx)/x-int(2dx)/(x-1)+ (7dx)/(x-1)^2#

#=-1/x+2lnx-2ln(x-1)-7/(x-1) +C#