Question

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

Answer 1

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`