Question Number 126907 by Dwaipayan Shikari last updated on 25/Dec/20
$$\boldsymbol{{Merry}}\:\boldsymbol{{christmas}}\:!! \\ $$$$ \\ $$π
π€ΆβοΈπππ¦
$$ \\ $$$$ \\ $$πππππππππ
ππππππππ
$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\boldsymbol{{tanh}}^{β\mathrm{1}} \boldsymbol{{x}}}{\:\sqrt[{\mathrm{5}}]{\boldsymbol{{x}}}}\boldsymbol{{dx}} \\ $$
$$ \\ $$$$ \\ $$πππππππππ
ππππππππ
$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\boldsymbol{{tanh}}^{β\mathrm{1}} \boldsymbol{{x}}}{\:\sqrt[{\mathrm{5}}]{\boldsymbol{{x}}}}\boldsymbol{{dx}} \\ $$
Answered by Olaf last updated on 25/Dec/20
$$\mathrm{The}\:\mathrm{function}\:\mathrm{tanh}^{β\mathrm{1}} \:\mathrm{is}\:\mathrm{defined} \\ $$$$\left.\mathrm{for}\:{x}\in\right]β\mathrm{1},+\mathrm{1}\left[.\:\right. \\ $$$$\mathrm{Not}\:\mathrm{for}\:{x}\in\left[\mathrm{1},{e}\right]…\mathrm{right}\:? \\ $$
Commented by Dwaipayan Shikari last updated on 25/Dec/20
$$\left.{I}\:{have}\:{corrected}\:{the}\:{question}\:\:\::\right) \\ $$
Commented by Dwaipayan Shikari last updated on 25/Dec/20
$${tanh}^{β\mathrm{1}} {x}=\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}β{x}}\right) \\ $$