Question-156933

Question Number 156933 by CAIMAN last updated on 17/Oct/21

Answered by mindispower last updated on 17/Oct/21

(1/t)=x ∫_(1/2) ^(1/3) ((ln(x))/(x(1+x)))dx=∫((ln(x))/x)−((ln(x))/(1+x))dx =((ln^2 (x))/2)−ln(x)ln(1+x)+∫((ln(1+x))/x)dx y=−x in ∫((ln(1+x))/x)dx =∫((ln(1−y))/y)dy=−Li_2 (y),Li_2 ...dilogarithm function ∫((ln(x))/(x(1+x)))=((ln^2 (x))/2)−ln(x)ln(1+x)−Li_2 (−x)+c ∫_(1/2) ^(1/3) ((ln(x))/(x(1+x)))dx=(1/2)(ln^2 (3)−ln^2 (3))−ln((1/3))ln((4/3))+ln((1/2))ln((3/2)) −li_2 (−(1/3))+li_2 (−(1/2))

$$\frac{\mathrm{1}}{{t}}={x} \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{3}}} \frac{{ln}\left({x}\right)}{{x}\left(\mathrm{1}+{x}\right)}{dx}=\int\frac{{ln}\left({x}\right)}{{x}}−\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$=\frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}}−{ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right)+\int\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$$${y}=−{x}\:{in}\:\int\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$$$=\int\frac{{ln}\left(\mathrm{1}−{y}\right)}{{y}}{dy}=−{Li}_{\mathrm{2}} \left({y}\right),{Li}_{\mathrm{2}} …{dilogarithm}\:{function} \\ $$$$\int\frac{{ln}\left({x}\right)}{{x}\left(\mathrm{1}+{x}\right)}=\frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}}−{ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right)−{Li}_{\mathrm{2}} \left(−{x}\right)+{c} \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{3}}} \frac{{ln}\left({x}\right)}{{x}\left(\mathrm{1}+{x}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}^{\mathrm{2}} \left(\mathrm{3}\right)−{ln}^{\mathrm{2}} \left(\mathrm{3}\right)\right)−{ln}\left(\frac{\mathrm{1}}{\mathrm{3}}\right){ln}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)+{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right){ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$−{li}_{\mathrm{2}} \left(−\frac{\mathrm{1}}{\mathrm{3}}\right)+{li}_{\mathrm{2}} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\: \\ $$$$ \\ $$$$ \\ $$

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