An excellent general result.

If \(\Re(s)>1\),
\[\displaystyle\int_0^\infty\frac{\ln x}{x^s+1}~\mathrm dx=\frac{\pi^2}{4s^2}\left[\sec^2\left(\frac{\pi}{2s}\right)-\csc^2\left(\frac{\pi}{2s}\right)\right]\]

Special cases:
\[\displaystyle\int_0^\infty\frac{\ln x}{x^2+1}~\mathrm dx=0\]
\[\displaystyle\int_0^\infty\frac{\ln x}{x^3+1}~\mathrm dx=-\frac{2\pi^2}{27}\]
\[\displaystyle\int_0^\infty\frac{\ln x}{x^4+1}~\mathrm dx=-\frac{\pi^2}{8\sqrt2}\]
\[\displaystyle\int_0^\infty\frac{\ln x}{x^5+1}~\mathrm dx=-\frac{4\pi^2}{25}\left(\frac{2+\sqrt5}{5+\sqrt5}\right)=-\frac{(5+3\sqrt5)\pi^2}{125}\]

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"How to differentiate parameters under the integral sign is a certain operation. It turns out that’s not taught very much in the universities; they don’t emphasize it." Richard Feynman's integral trick 👉 🔗 https://www.cantorsparadise.com/richard-feynmans-integral-trick-e7afae85e25c

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Riemann-Integral vs. Lebesgue-Integral:

https://friendica.a-zwenkau.de/display/fdacc1ed-2163-2ea4-3c06-4f0352187840