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Calculus Level 5

I=∫0∞xln⁡(4x)14x4+x2+1 dx\large I = \int_{0}^{\infty}\dfrac{x \ln{(4x)}}{\frac{1}{4}x^{4}+x^{2}+1} \,dx

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

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

Brandon Monsen by Brandon Monsen
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