HP Prime integration issue

[Albert Chan]

Hi, LCieParagon

I asked this before, and you are not supposed to see rootof(...) in Prime.
see https://www.hpmuseum.org/forum/thread-11...#pid105773

For your example, we can do a simple substitution, x = 2^(1/3)*y, dx = 2^(1/3)*dy
(y integral result copy/pasted from Hp Prime emulator)

I = ∫(1/(x^3-2), x)
  = 2^(1/3)/2 * ∫(1/(y^3-1), y)
  = 2^(1/3)/2 * (-atan((y+1/2)/(1/2*√3))/√3 - 1/6*ln(y^2+y+1) + 1/3*ln(abs(y-1)))

Back substitute y = x/2^(1/3), and we are done.

---

The reason for substitution is y^3-1 = (y-1)*(y-w)*(y-w²), where w = exp(i*2*pi/3)

3/(y^3-1) = 1/(y-1) + w/(y-w) + w²/(y-w²)

I = 2^(1/3)/6 · ∫(3/(y^3-1), y)
  = 2^(1/3)/6 · (ln(y-1) + w·ln(y-w) + w²·ln(y-w²))