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Question Number 17512 by tawa tawa last updated on 06/Jul/17

$$\int \frac{{\mathrm{e}}^{-\mathrm{t}}\ln (1+{\mathrm{e}}^{-\mathrm{t}})}{1+{\mathrm{e}}^{-\mathrm{t}}}\mathrm{dt}$$

Answered by sma3l2996 last updated on 07/Jul/17

$$I=\int \frac{e^{-t}ln(1+e^{-t})}{1+e^{-t}}dt=\int {(-ln(1+e^{-t}))}^{\prime }ln(1+e^{-t})$$$$=-\frac{1}{2}{\left(ln(1+e^{-t})\right)}^2+C$$

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