Question

evaluate the integral:

integral of 13/(x^3 - 27) dx

Answer 1

use partial fractions to get

13/27 (1/(x-3) - (x+6)/(x^2+3x+9))
13/27 (1/(x-3) - 1/2 (2x+3)/(x^2+3x+9) - (9/2)/(x^2+3x+9))

13/27 (1/(x-3) - 1/2 (2x+3)/(x^2+3x+9) - (9/2)/(x+3/2)^2 + 27/4)

Now you can use substitutions
u = x-3
v = x^2+3x+9
w = x + 3/2
to come up with

13/27 (du/u - 1/2 dv/v - 9/2 / (w^2+(√27/2)^2))

Then using the trig substitution for
∫ 1/(w^2+a^2) = 1/a arctan(w/a)

everything falls right out.