Hi, I think your numerical integration is correct here.

Using Maple for expressing x as function of y by quadrature, I obtain

x(y) = ln(1+y)-1/2*ln(y^2-y+1)+3^(1/2)*arctan(1/3*3^(1/2)*(2*y-1))+1/6*3^(1/2)*Pi

The limit of x for y -> +\infty is 2/3*3^(1/2)*Pi, approximate

3.627598730.

See the maple output below. I also verified that x(0)=0 and dx/dy * (1+y^3)/3 =1 for the solution given above.

Best regards

Herbert

> exactly:=x=int(3/(1+v^3),v=0..y);X:=unapply(simplify(rhs(exactly),ln),y);X1:=limit(X(Y),Y=infinity);evalf(X1);

exactly := x = ln(1+y)-1/2*ln(y^2-y+1)+3^(1/2)*arctan(1/3*3^(1/2)*(2*y-1))+1/6*3^(1/2)*Pi

X := proc (y) options operator, arrow; ln(y+1)-1/2*ln(y^2-y+1)+3^(1/2)*arctan(1/3*3^(1/2)*(2*y-1))+1/6*3^(1/2)*Pi end

X1 := 2/3*3^(1/2)*Pi

3.627598730

Thanks a lot Herbert.

I will do again the formal part.

Regards,

Gabriel

Thanks @Herbert

 Hi. We  have an equation with separable variables. So we can integrate it in quadratures and elemental functions which was made by prof.  Herbert H H Homeier. ·Then it is simple to calculate limit of x(y)  for y \to +\infty by using the L' opital's rule.

I agree with Herbert, see also my visual contribution.