Indefinite integral of x^2/(x^6+9)

My hint is to use arctan but because my teacher didn't explain the inverse trig function very well, I don't know what I'm doing. Please help!

Note that ∫ x^2 dx/(x^6 + 9) = ∫ x^2 dx/((x^3)^2 + 9).

So, let u = x^3, du = 3x^2 dx.

Hence, the integral becomes
∫ (1/3) du/(u^2 + 9)
= (1/3) * (1/3) arctan(x/3) + C
= (1/9) arctan(x^3/3) + C.

I hope this helps!