Integration Questions

Question Number 161656 by amin96 last updated on 20/Dec/21

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{{xln}}\left(\mathrm{1}+\boldsymbol{{x}}\right)}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{dx}} \\$$

Commented by Ar Brandon last updated on 21/Dec/21

Commented by amin96 last updated on 21/Dec/21

$${solution}??? \\$$

Commented by smallEinstein last updated on 22/Dec/21

Answered by mindispower last updated on 21/Dec/21

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{ln}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right){dx}={A} \\$$$${y}=\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\Rightarrow{x}=\frac{\mathrm{1}−{y}}{\mathrm{1}+{y}}\Rightarrow{dx}=\frac{−\mathrm{2}{dy}}{\left(\mathrm{1}+{y}\right)^{\mathrm{2}} } \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{y}}{\mathrm{1}+{y}}.\frac{{ln}\left({y}\right)}{\mathrm{1}+{y}^{\mathrm{2}} }.{dy} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({y}\right)}{\mathrm{1}+{y}}−\frac{{yln}\left({y}\right)}{\mathrm{1}+{y}^{\mathrm{2}} }{dy}=\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({y}\right)}{\mathrm{1}+{y}}{dy} \\$$$$=−\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{y}\right)}{{y}}{dy}=\frac{\mathrm{3}}{\mathrm{4}}\left(−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−\left(−{y}\right)\right)}{\left(−{y}\right)}{d}\left(−{y}\right)\right) \\$$$$=\frac{\mathrm{3}}{\mathrm{4}}{Li}_{\mathrm{2}} \left(−\mathrm{1}\right)=\frac{\mathrm{3}}{\mathrm{4}}.−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{y}\right)}{\mathrm{1}+{y}}{dt}={B} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{\mathrm{2}−{t}}{dt} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\frac{{t}}{\mathrm{2}}\right)+{ln}\left(\mathrm{2}\right)}{\mathrm{1}−\left(\frac{{t}}{\mathrm{2}}\right)}{d}\left(\frac{{t}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{ln}\left({u}\right)}{\mathrm{1}−{u}}{du}+\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{du}}{\mathrm{1}−{u}} \\$$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{ln}\left(\mathrm{1}−\left(\mathrm{1}−{u}\right)\right)}{\mathrm{1}−{u}}{d}\left(\mathrm{1}−{u}\right) \\$$$$\left.=\frac{\mathrm{1}}{\mathrm{2}}{Li}_{\mathrm{2}} \left(\mathrm{1}−{u}\right)\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} −\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−{Li}_{\mathrm{2}} \left(\mathrm{1}\right)+{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\right] \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}\right]=−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}+\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{4}} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{ln}\left(\mathrm{1}+{x}\right){dx}=\frac{{A}+{B}}{\mathrm{2}}=−\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{96}}+\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{8}} \\$$$$= \\$$$$\\$$

Commented by smallEinstein last updated on 21/Dec/21

pls recheck solution