Integral of (√x dx)/(x-∛x)

∫(√x dx)/(x-∛x)

Any ideas would be greatly appreciated because I have none

int(x^(1/2) (x-(x^(1/3))^-1))
integration by parts
let
u=x^(1/2), u'=(1/2)x^(-1/2)
dv=x-x^(1/3)
v=x^2/2 - (3/4)(x^(4/3))
uv-int(v, du)=after computations,
-2/5 x^(5/2) - (3/4)x^(11/6) + (15/32) x^(5/4) +k

You can say x=y^6.
Then you don't have any root left.

After that, you can divide the numerator by the denominator.

You get an immediate integral and a integral which can be resolved using the partial fractions method.

Then, put back x in your answer.

You should get something with a square root of x,a few logs (probably two) and a tan-1.

x = u^3
dx = 3u^2 du

Integral of u^(2/3) * 3u^2 du / (u^3 - u)

Integral of u^(8/3)du / (u(u^2-1))

Integral of u^(5/3)du / (u^2 - 1)

Let u = sec w
du = sec w tan w dw

Integral of sec^(5/3) w * sec w tan w dw / (sec^2 w - 1)

Integral of sec^(5/3) w * sec w tan w dw / (tan^2 w)