Question Number 205264 by pticantor last updated on 13/Mar/24
$$\boldsymbol{{pls}}\:\boldsymbol{{how}}\:\boldsymbol{{to}}\:\boldsymbol{{calculate}}\:\boldsymbol{{this}}? \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\boldsymbol{{ln}}\left(\boldsymbol{{x}}+\mathrm{1}\right)}{\boldsymbol{{x}}}\boldsymbol{{dx}} \\ $$
Answered by Berbere last updated on 13/Mar/24
![=−(−∫_(1/2) ^1 ((ln(1−(−x)))/((−x)))d(−x)) =−[Li_2 (−x)]_(1/2) ^1 =−Li_2 (−1)+Li_2 (−(1/2)) =(π^2 /(12))+Li_2 (−(1/2))](Q205267.png)
$$=−\left(−\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−\left(−{x}\right)\right)}{\left(−{x}\right)}{d}\left(−{x}\right)\right) \\ $$$$=−\left[{Li}_{\mathrm{2}} \left(−{x}\right)\right]_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} =−{Li}_{\mathrm{2}} \left(−\mathrm{1}\right)+{Li}_{\mathrm{2}} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+{Li}_{\mathrm{2}} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$