Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 17

int(x^2-1)/(x^3+x)dx
(x^2-1)/(x^3+x)=(x^2-1)/(x(x^2+1))
Now let's create partial fraction template,
(x^2-1)/(x(x^2+1))=A/x+(Bx+C)/(x^2+1)
Multiply equation by the denominator,
(x^2-1)=A(x^2+1)+(Bx+C)x
(x^2-1)=Ax^2+A+Bx^2+Cx
x^2-1=(A+B)x^2+Cx+A
Comparing the coefficients of the like terms,
A+B=1 ----------------(1)
C=0
A=-1
Plug the value of A in equation 1,
-1+B=1
B=2
Plug in the values of A,B and C in the partial fraction template,
(x^2-1)/(x(x^2+1))=-1/x+(2x)/(x^2+1)
int(x^2-1)/(x^3+x)dx=int(-1/x+(2x)/(x^2+1))dx
Apply the sum rule,
=int-1/xdx+int(2x)/(x^2+1)dx
Take the constant out,
=-1int1/xdx+2intx/(x^2+1)dx
Now evaluate both the integrals separately,
int1/xdx=ln|x|
Now let's evaluate second integral,
intx/(x^2+1)dx
Apply integral substitution: u=x^2+1
du=2xdx
=int1/u(du)/2
=1/2int1/udu
=1/2ln|u|
Substitute back u=x^2+1
=1/2ln|x^2+1|
int(x^2-1)/(x^3+x)dx=-ln|x|+2(1/2ln|x^2+1|)
Simplify and add a constant C to the solution,
=-ln|x|+ln|x^2+1|+C