Okay, I figured it out. It's a little tricky because you use both integration by parts and substitution.
Since I don't want to deal with constants, I'll drop the 4/9 in my calculations.
Start by doing integration by parts with u=(1+x^(-2/3))^(1/2) and dv=dx.
Then du=(-5/6)x^(-5/3)(1+x^(-2/3))^(-1/2)dx and v=x.
Now our original integral is equal to uv-int vdu, so I'll concentrate on the integral of vdu.
-vdu=(5/6)x^(-2/3)(1+x^(-2/3))^(-1/2)dx
This looks bad, but put all negative powers in the denominator and distribute half of the x power (x^(1/3)) into the square root. This multiplies the terms inside the square root by x^(2/3) and we are left with:
(5/6)1/(x^(1/3)(x^(2/3)+1)^1/2))
Do a substitution of u=1+x^(2/3). Then du=x^(-1/3), which is exactly the power of x we have outside the square root.
To finish, you just have int(u^(-1/2))du, which is an easy power rule, and then back substitution for x.
Hope this helps!