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Question Number 87861 by Rio Michael last updated on 06/Apr/20

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\:\mathrm{tanh}\:\mathrm{2}{x}\:{dx} \\$$

Commented by MJS last updated on 06/Apr/20

$$\mathrm{tanh}\:\mathrm{2}{x}\:=\frac{\mathrm{e}^{\mathrm{4}{x}} −\mathrm{1}}{\mathrm{e}^{\mathrm{4}{x}} +\mathrm{1}} \\$$$$\int\mathrm{tanh}\:\mathrm{2}{x}\:{dx}= \\$$$$\:\:\:\:\:\left[{t}=\mathrm{e}^{\mathrm{4}{x}} \:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{4e}^{\mathrm{4}{x}} }\right] \\$$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{t}−\mathrm{1}}{{t}\left({t}+\mathrm{1}\right)}{dt}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dt}}{{t}}= \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({t}+\mathrm{1}\right)\:−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:{t}\:=−{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{e}^{\mathrm{4}{x}} +\mathrm{1}\right)\:+{C} \\$$

Commented by Rio Michael last updated on 06/Apr/20

$$\:\mathrm{sir}\:\mathrm{can}\:\mathrm{we}\:\mathrm{use}\:''\:\boldsymbol{\mathrm{integration}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{parts}}''? \\$$

Commented by MJS last updated on 07/Apr/20

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{that}\:\mathrm{would}\:\mathrm{be}\:\mathrm{easier} \\$$

Commented by Rio Michael last updated on 07/Apr/20

$$\mathrm{i}\:\mathrm{will}\:\mathrm{try}\:\mathrm{it}\:\mathrm{out}\:\mathrm{sir} \\$$