Integration Questions

Question Number 176542 by mnjuly1970 last updated on 20/Sep/22

$$\\$$$$\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{x}.{tanh}^{\:−\mathrm{1}} \left({x}\right)}{\left(\mathrm{1}+{x}\right)^{\:\mathrm{2}} }{dx}=\:\frac{\mathrm{1}}{\mathrm{24}}\:\left(\pi^{\:\mathrm{2}} −\mathrm{6}\right) \\$$

Answered by Peace last updated on 20/Sep/22

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{tg}^{−} \left({x}\right){dx} \\$$$${tanh}^{−} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}−{x}\right)\right) \\$$$$\frac{{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{1}+{x}}−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} } \\$$$$\mathrm{2}\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}−\frac{{ln}\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}{dx} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\left[\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }+{A}+{B} \\$$$$=\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}+\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}+{A}+{B} \\$$$${A}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\int_{\mathrm{0}} ^{{x}} \frac{{ln}\left(\mathrm{1}−{t}\right)}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left\{−\frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}−\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }.{dt}\right\} \\$$$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}−\frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{1}}{\mathrm{1}−{t}}+\frac{\mathrm{1}}{\mathrm{1}+{t}}{dt} \\$$$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}−\frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}+\frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{2}}−\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{2}} \\$$$$=−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\$$$$−{B}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}{dx} \\$$$${x}=\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}},=−\mathrm{1}+\frac{\mathrm{2}}{\mathrm{1}+{t}}\Rightarrow{dt}=−\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} } \\$$$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}}\right)}{\frac{\mathrm{2}}{\left(\mathrm{1}+{t}\right)}}.\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{2}{t}\right)−{ln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}}{dt} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{2}\right)}{\mathrm{1}+{t}}+\frac{{ln}\left({t}\right)}{\mathrm{1}+{t}}−\frac{{ln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}}{dt} \\$$$$={ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\boldsymbol{{t}}\right)}{\mathrm{1}+{t}}{dt}=\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−\left(−{t}\right)\right.}{−{t}}{d}\left(−{t}\right) \\$$$$=\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}−{Li}_{\mathrm{2}} \left(−\mathrm{1}\right) \\$$$${B}=−\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}+{Li}_{\mathrm{2}} \left(−\mathrm{1}\right) \\$$$$\mathrm{2}\Omega=\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}+\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}+{A}+{B} \\$$$$=\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}+\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}−\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}+{Li}_{\mathrm{2}} \left(−\mathrm{1}\right) \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}+{Li}_{\mathrm{2}} \left(−\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{12}}\left(\pi^{\mathrm{2}} −\mathrm{6}\right) \\$$$$\Omega=\frac{\mathrm{1}}{\mathrm{24}}\left(\pi^{\mathrm{2}} −\mathrm{6}\right) \\$$$$\\$$$$\\$$$$\\$$

Commented by Tawa11 last updated on 22/Sep/22

$$\mathrm{Great}\:\mathrm{sir} \\$$

Commented by Peace last updated on 23/Sep/22

$${thank}\:{You}\:{nice}\:{Day} \\$$