Integration Grandmaster - P7

With the substitution $\sin \varphi \; = \; \frac{x}{\sqrt{x^3 + 1}} \hspace{2cm} 0 \le x \le 1$ we see that $\begin{aligned} \cos\varphi\frac{d\varphi}{dx} & = \; \frac{1}{\sqrt{x^3+1}} - \frac{3x^3}{2(x^3+1)\sqrt{x^3+1}} \; = \; \frac{2 - x^3}{2(x^3+1)\sqrt{x^3+1}} \\ \cos\varphi & = \; \frac{\sqrt{x^3 - x^2 +1}}{\sqrt{x^3+1}} \end{aligned}$ so that $A \; = \; \int_0^1 \frac{(x^3-2)\sqrt{x^3 - x^2 + 1}}{(x^3+1)^2}\,dx \; = \; -2\int_0^{\frac14\pi}\cos^2\varphi\,d\varphi \; = \; -\Big[\varphi + \tfrac12\sin2\varphi\Big]_0^{\frac14\pi} \; = \; -\tfrac14\pi - \tfrac12$ and hence $\lfloor 10000 A \rfloor = \boxed{-12854}$ .