An integral for $\frac{9801}{2206\sqrt{2}}-\pi$ From integrals
$$\pi=\frac{24\sqrt{2}}{11} + \frac{8}{11} \int_0^1 \frac{x (1 - x)^2(1 + 2 \sqrt{2} x^4)}{1 + x^2 + x^4 + x^6} dx$$
and
$$\pi=\frac{20\sqrt{2}}{9} - \frac{2\sqrt{2}}{3} \int_0^1 \frac{x^4(1 - x)^4}{1 + x^2 + x^4 + x^6} dx$$
the following linear combination for Ramanujan's approximation $\pi\approx\frac{9801}{2206\sqrt{2}}$  is obtained:
$$\pi=\frac{9801}{2206\sqrt{2}} - \frac{1}{8824}\int_0^1 \frac{x (1 - x)^2 (124 (1 + 2 \sqrt{2} x^4) - 5769 \sqrt{2} x^3 (1 - x)^2)}{1 + x^2 + x^4 + x^6} dx$$
This integrand is small but has sign changes in $\left(0,1\right)$, so it does not provide a direct proof that $\pi<\frac{9801}{2206\sqrt{2}}$ such as Dalzell integral for $\pi<\frac{22}{7}$.
$$\pi = \frac{22}{7}-\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx$$

Is there an integral for $\frac{9801}{2206\sqrt{2}}-\pi$ with positive integrand? 

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HINT: write your Integrand in the form
$$2\,\sqrt {2}x-4\,\sqrt {2}+{\frac {2\,\sqrt {2}+1}{{x}^{2}+1}}+{\frac 
{2\,{x}^{2}\sqrt {2}-2\,\sqrt {2}x-{x}^{2}+2\,\sqrt {2}+x-1}{{x}^{4}+1
}}
$$
