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Question Number 101597 by Rio Michael last updated on 03/Jul/20

 ∫ ln (1+ e^x ) dx = ..

$$\:\int\:\mathrm{ln}\:\left(\mathrm{1}+\:{e}^{{x}} \right)\:{dx}\:=\:.. \\ $$

Commented by Dwaipayan Shikari last updated on 03/Jul/20

∫(e^x −(1/2)e^(2x) +(1/3)e^(3x) −(1/4)e^(4x) +(1/5)e^(5x) −(1/6)e^(6x) +....)dx  e^x −(e^(2x) /4)+(e^(3x) /9)−(e^(4x) /(16))+(e^(5x) /(25))−(e^(6x) /(36))+(e^(7x) /(49))−(e^(8x) /(64))+.......  It is approximate

$$\int\left({e}^{{x}} −\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{x}} +\frac{\mathrm{1}}{\mathrm{3}}{e}^{\mathrm{3}{x}} −\frac{\mathrm{1}}{\mathrm{4}}{e}^{\mathrm{4}{x}} +\frac{\mathrm{1}}{\mathrm{5}}{e}^{\mathrm{5}{x}} −\frac{\mathrm{1}}{\mathrm{6}}{e}^{\mathrm{6}{x}} +....\right){dx} \\ $$$${e}^{{x}} −\frac{{e}^{\mathrm{2}{x}} }{\mathrm{4}}+\frac{{e}^{\mathrm{3}{x}} }{\mathrm{9}}−\frac{{e}^{\mathrm{4}{x}} }{\mathrm{16}}+\frac{{e}^{\mathrm{5}{x}} }{\mathrm{25}}−\frac{{e}^{\mathrm{6}{x}} }{\mathrm{36}}+\frac{{e}^{\mathrm{7}{x}} }{\mathrm{49}}−\frac{{e}^{\mathrm{8}{x}} }{\mathrm{64}}+....... \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{approximate} \\ $$

Commented by floor(10²Eta[1]) last updated on 03/Jul/20

Σ_(k=1) ^∞ (((−1)^(k+1) e^(kx) )/k^2 )+C

$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}+\mathrm{1}} \mathrm{e}^{\mathrm{kx}} }{\mathrm{k}^{\mathrm{2}} }+\mathrm{C} \\ $$

Commented by Rio Michael last updated on 03/Jul/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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