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Question Number 93553 by Ar Brandon last updated on 13/May/20

$$\int ln(1+{\mathrm{e}}^{\mathrm{u}})\mathrm{du}$$

Commented by Tony Lin last updated on 13/May/20

$$let-e^u=zdz=zdu$$$$\int \frac{ln(1-z)}{z}dz$$$$=-{Li}_2\left(z\right)+c$$$$=-{Li}_2(-e^u)+c$$$${Li}_2\left(z\right)=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{z^k}{k^2}$$

Commented by Ar Brandon last updated on 13/May/20

Okay thanks ��

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