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

1 Answer
Narad T.
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

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