The slope of the tangent to the curve $y=\int_0^x\frac{dt}{1+t^3}$ at the point where x = 1 is:

1/2 1 1/4 None

1/2

$\frac{dy}{dx}=\frac{1}{(1+x^3)}\frac{dy}{dx}=\frac{1}{1+x^3}(1)=\frac{1}{x^3+1}=\frac{1}{1^3+1}=\frac{1}{2}$