Question Number 17512 by tawa tawa last updated on 06/Jul/17
$$\int\:\frac{\mathrm{e}^{−\mathrm{t}} \:\mathrm{ln}\left(\mathrm{1}\:+\:\mathrm{e}^{−\mathrm{t}} \right)}{\mathrm{1}\:+\:\mathrm{e}^{−\mathrm{t}} }\:\mathrm{dt} \\ $$
Answered by sma3l2996 last updated on 07/Jul/17
$${I}=\int\frac{{e}^{−{t}} {ln}\left(\mathrm{1}+{e}^{−{t}} \right)}{\mathrm{1}+{e}^{−{t}} }{dt}=\int\left(−{ln}\left(\mathrm{1}+{e}^{−{t}} \right)\right)'{ln}\left(\mathrm{1}+{e}^{−{t}} \right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{1}+{e}^{−{t}} \right)\right)^{\mathrm{2}} +{C} \\ $$