Question Number 175053 by mnjuly1970 last updated on 17/Aug/22 | ||
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Commented by infinityaction last updated on 18/Aug/22 | ||
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$$\sqrt{\mathrm{2}}{tanh}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$ | ||
Commented by Frix last updated on 18/Aug/22 | ||
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$$=\sqrt{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$ | ||