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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} \,dxI=∫0∞41x4+x2+1xln(4x)dx
III is the value of the closed form of the above integral. What is the value of ⌊100I⌋\lfloor 100I \rfloor⌊100I⌋?
You may use the fact that ln(2)≈0.693147\ln(2) \approx 0.693147ln(2)≈0.693147.
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