UNKNOWN Questions

Question Number 59049 by ugwu Kingsley last updated on 03/May/19

$$\:\underset{\:\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{x}\:\mathrm{log}\:{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:= \\$$

Commented bymaxmathsup by imad last updated on 04/May/19

$${let}\:\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{xln}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\Rightarrow\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{xln}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}+\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{xln}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx} \\$$ $$\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{xln}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:=_{{x}=\frac{\mathrm{1}}{{t}}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{{t}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:{ln}\left(\frac{\mathrm{1}}{{t}}\right)\frac{−{dt}}{{t}^{\mathrm{2}} } \\$$ $$=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{ln}\left({t}\right)}{{t}^{\mathrm{3}} \left(\frac{\mathrm{1}+{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:{dt}\:\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{tln}\left({t}\right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow\:{A}\:=\mathrm{0}\:. \\$$