Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 24

Integrate int(x^2-x+6)/(x^3+3x)
Rewrite the rational function using partial fractions.
(x^2-x+6)/(x^3+3x)=(A/x)+(Bx+C)/(x^2+3)
x^2-x+6=A(x^2+3)+(Bx+C)x
x^2-x+6=Ax^2+3A+Bx^2+Cx
x^2-x+6=(A+B)x^2+Cx+3A
Equate coefficients and solve for A, B, and C.
3A=6
A=2

C=-1

A+B=1
2+B=1
B=-1

int(x^2-x+6)/(x^3+3x)dx=int(2/x)dx+int[(x-1)/(x^2+3)]dx
=int(2/x)dx+int[x/(x^2+3)]dx-int[1/(x^2+3)]dx

The first integral matches the form int (du)/u=ln|u|+C
int(2/x)dx=2int(1/x)dx=2ln|x|+C

Integrate the second integral using u-substitution.
Let u=x^2+3
(du)/(dx)=2x
dx=(du)/(2x)
-intx/(x^2+3)dx
=-int(x/u)*(du)/(2x)
=-1/2ln|u|
=-1/2ln|x^2+3|

The third integral matches the form
int(dx)/(x^2+a^2)=(1/a)tan^-1(x/a)+C
-int1/(x^2+3)dx
=-1/sqrt3tan^-1(x/sqrt3)+C

The final answer is:
2ln|x|-1/2ln|x^2+3|-1/sqrt3tan^-1(x/sqrt3)+C