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#
