\[\large \displaystyle\int_0^{\infty}\frac{x^4e^x}{(e^x-1)^2} \, dx\]

The integral above equals to \( \frac ab \pi^4 \) for coprime positive integers \(a,b\) and that you're given \( \displaystyle \sum_{j=1}^\infty \frac1{j^4} = \frac{\pi^4}{90} \).

Find \(a+b\)

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