Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 21

Integrate int(x^3+4)/(x^2+4)dx
Rewrite the given function using long division.
int[x+(-4x+4)/(x^2+4)]dx
=intxdx-int(4x)/(x^2+4)dx+int4/(x^2+4)dx

Integrate the first integral using the pattern intx^n=x^(n+1)/n+C
intx=x^2/2+C

Integrate the second integral using u-substitution.
let u=x^2+4
(du)/dx=2x
dx=(du)/(2x)
-int(4x)/(x^2+4)dx
=-4intx/u*(du)/(2x)

=-2ln|x^2+4|+C
=-2ln(x^2+4)+C

Integrate the third integral using the pattern
int(dx)/(x^2+a^2)=(1/a)tan^-1(x/a)+C
int4/(x^2+4)dx=(4)(1/2)tan^-1(x/2)+C=2tan^-1(x/2)+C

The final answer is:
1/2x^2-2ln(x^2+4)+2tan^-1(x/2)+C