Use long division on the integrand:
#color(white)( (x^2+36)/color(black)(x^2+36))color(black)(x" ")/(")" color(white)(x)x^3+0x^2+0x+36)#
#" "color(black)(-x^3-0x^2-36x" ")/(" "-36x+36)#
Written another way:
#(x^3+36)/(x^2+36)=x-(36x)/(x^2+36) + 36/(x^2+36)#
Check:
#(x(x^2+36))/(x^2+36)-(36x)/(x^2+36)+36/(x^2+36)#
#(x^3+36x)/(x^2+36)-(36x)/(x^2+36)+36/(x^2+36)#
#(x^3+36)/(x^2+36)#
This checks.
The original integral becomes 3 integrals:
#int(x^3+36)/(x^2+36)dx=intxdx - 18int(2x)/(x^2+36)dx+36int1/(x^2+36)dx#
The first integral is trivial:
#int(x^3+36)/(x^2+36)dx=x^2/2 - 18int(2x)/(x^2+36)dx+36int1/(x^2+36)dx#
I deliberately set up the second integral for a variable substitution resulting in the natural logarithm:
#int(x^3+36)/(x^2+36)dx=x^2/2 - 18ln(x^2+36)+36int1/(x^2+36)dx#
The third integral is our old friend the inverse tangent:
#int(x^3+36)/(x^2+36)dx=x^2/2 - 18ln(x^2+36)+6tan^-1(x/6)dx+C#
