My approach to this type of integral is to recognise that 3 x^2 is the

derivative of x^3+3, so this suggest the integral is of the form:

integral (x^2) / sqrt (x^3 + 3) ) dx = k sqrt(x^3 + 3)

then we differentiate the right hand side and equate that to the integrand

to determine k.

(d/dx) k sqrt(x^3 + 3) = k (3x^2) (1/2) /sqrt(x^3+3)

......................=(3k/2) [x^2/sqrt(x^3+3)]

So we need 3k/2=1, or k=2/3, hence:

integral (x^2) / sqrt (x^3 + 3) ) dx = (2/3) sqrt(x^3 + 3)

RonL