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Question Number 117841 by john santu last updated on 14/Oct/20

∫ ln (1−e^(−2x) ) dx =?

$$\int\:\mathrm{ln}\:\left(\mathrm{1}−{e}^{−\mathrm{2}{x}} \right)\:{dx}\:=? \\ $$

Answered by MJS_new last updated on 14/Oct/20

∫ln (1−e^(−2x) ) dx=       [t=e^(−2x)  → dx=−(e^(2x) /2)dt]  =−(1/2)∫((ln (1−t))/t)dt=(1/2)Li_2  t =(1/2)Li_2  e^(−2x)  +C

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

Commented by john santu last updated on 14/Oct/20

thank prof

$${thank}\:{prof} \\ $$

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