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$\large I = \int_{0}^{\infty}\dfrac{x \ln{(4x)}}{\frac{1}{4}x^{4}+x^{2}+1} \,dx$

$I$ is the value of the closed form of the above integral. What is the value of $\lfloor 100I \rfloor$?

You may use the fact that $\ln(2) \approx 0.693147$.

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