Question Number 76664 by aliesam last updated on 29/Dec/19
$$\int_{\mathrm{0}} ^{+\infty} \frac{{t}\:{ln}\left({t}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:{dt} \\ $$
Commented by mathmax by abdo last updated on 29/Dec/19
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{tln}\left({t}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\Rightarrow{I}\:=_{{t}=\frac{\mathrm{1}}{{x}}} \:\:−\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{−{ln}\left({x}\right)}{{x}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}\right)^{\mathrm{2}} }\left(−\frac{{dx}}{{x}^{\mathrm{2}} }\right) \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left({x}\right)}{{x}^{\mathrm{3}} ×\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{x}^{\mathrm{4}} }}{dx}\:=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{xln}\left({x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\Rightarrow\mathrm{2}{I}\:=\mathrm{0}\:\Rightarrow\:{I}\:=\mathrm{0} \\ $$