Ouch, that is a tough one!

But I didn't get around to solving "Student"s integral. Which I will attempt now ...

∫ x^2/(1 - 2x^3) dx

Try: u=1-2x^3

Then: du = -6x^2 dx (ie we differentiated it)

So: x^2 dx = (-1/6) du

Substituting **1-2x^3=u** and **x^2 dx = (-1/6) du** we get:

∫ x^2/(1 - 2x^3) dx = ∫ 1/u (-1/6) du

Then it is fairly easy to solve ∫ 1/u (-1/6) du:

∫ 1/u (-1/6) du = (-1/6) ∫ 1/u du = (-1/6) ln u

Now put back u=1-2x^3:

(-1/6) ln( 1-2x^3 )

DONE! (I hope!)