Can someone tell me how $$−56−173\,\ln(11)+366\,\ln(13)−\left(\frac{105}2+20\,\ln(2)+366\,\ln(3)\right)$$ simplifies to $$\frac{-217}2−20\,\ln(2)−173\,\ln(11)+732\,{\rm arctanh}\left(\frac58\right)?$$
I just don't know how to get to the arctan part.
2026-03-29 02:34:39.1774751679
can someone explain this simplification for me??
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$\tanh^{-1}(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$
$366\ln(13)-366\ln(3)=\frac{1}{2}(732ln(13)-732ln(3))=\frac{1}{2}\left(732\ln\left(\frac{13}{3}\right)\right)=\frac{1}{2}\left(732\ln\left(\frac{1+\frac{5}{8}}{1-\frac{5}{8}}\right)\right)=732\tanh^{-1}(\frac{5}{8})$