No, it's a simple substitution.
[tex]u = x^3 + 1[/tex]
Differentiating both sides with respect to x:
[tex]\frac{du}{dx} = 3x^2[/tex]
or equivalently
[tex]du = 3x^2 dx[/tex]
Now substitute into the original integral. The [itex]3x^2 dx[/itex] turns into [itex]du[/itex], and [itex]\sqrt{x^3 + 1}[/itex] becomes [itex]\sqrt{u}[/itex]. The endpoints of the integral also change accordingly.