Let's decompose 1/(x^2(x-1)) with partial fractions:
1/(x^2(x-1))=A/x+B/x^2+C/(x-1) where A, B, and C are real values we must find. We decompose in this manner because we have linear factors in the denominator.
Add up the second side:
1/(x^2(x-1))=(Ax(x-1))/(x^2(x-1))+(B(x-1))/(x^2(x-1))+(Cx^2)/(x^2(x-1))
We may now set numerators equal:
1=Ax(x-1)+B(x-1)+Cx^2
Let's find A, B, C:
Set x=0, thereby eliminating all terms including A, C:
1=-B, B=-1
Set x=1:
1=C(1^2), C=1
We can't solve for A in this manner, so let's multiply out the right side:
1=Ax^2-Ax+Bx-B+Cx^2
Group and factor terms containing the same degree of x:
1=Ax^2+Cx^2+Bx-A-B
1=x^2(A+C)+x(B-A)-B
A+C=0 as there is no term including x^2 on the left side, so the sum of A, C must be zero.
A+C=0
A+1=0, A=-1
So,
intdx/(x^2(x-1))dx=int(-1/x-1/x^2+1/(x-1))dx
Split up the integral:
intdx/(x^2(x-1))dx=int-1/xdx-int1/x^2dx+int1/(x-1)dx
Integrate:
intdx/(x^2(x-1))=-ln|x|+1/x+ln|x-1|+C
Simplify by combining logs:
intdx/(x^2(x-1))=ln|(x-1)/x|+1/x+C