Integral of 1/x^2*(x-1) dx =?

1 Answer
Mar 21, 2018

intdx/(x^2(x-1))=ln|(x-1)/x|+1/x+C

Explanation:

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