Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 18

Integrate int(x^2+2x-1)/(x^3-x)dx
Rewrite the rational function using partial fractions.
(x^2+2x-1)/(x^3-x)=A/x+B/(x+1)+C/(x-1)
x^2+2x-1=A(x^2-1)+Bx(x-1)+Cx(x+1)
x^2+2x-1=Ax^2-A+Bx^2-Bx+Cx^2+Cx
x^2+2x-1=(A+B+C)x^2+(C-B)x-A
Equate coefficients and solve for A, B, and, C.
-A=-1
A=1

A+B+C=1
1+B+C=1
B+C=0

C-B=2
C+B=0
2C=2
C=1

B+C=0
B+1=0
B=-1

int(x^2+2x-1)/(x^3-x)dx=int(1/x)dx-int1/(x+1)dx+1/(x-1)dx
=ln|x|-ln|x+1|+ln|x-1|+C
=ln|[x(x-1)]/(x+1)|+C

The final answer is:
=ln|[x(x-1)]/(x+1)|+C