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∫01ln(x4−2x2+5)−ln(5x4−2x2+1)1−x2 dx {\int_0^1\frac{\ln\left(x^4-2x^2+5\right)-\ln\left(5x^4-2x^2+1\right)}{1-x^2}\, dx} ∫011−x2ln(x4−2x2+5)−ln(5x4−2x2+1)dx
Given that the integral above is equal to παarctanβ−γδ \pi^\alpha\arctan\sqrt{\dfrac{\sqrt{\beta}-\gamma}{\delta}} παarctanδβ−γ, where α,β,γ,δ\alpha, \beta, \gamma, \delta α,β,γ,δ are integers with β\betaβ square-free.
Calculate α+β+γ+δ{\alpha+\beta+\gamma+\delta}α+β+γ+δ.
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