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