Question Number 172064 by qaz last updated on 23/Jun/22
$${Calculate}\:\:::\:\:\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\frac{\int_{\mathrm{1}} ^{{x}} \frac{{lnt}}{\mathrm{1}+{t}}{dt}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }=? \\ $$
Answered by puissant last updated on 23/Jun/22
$$=\:\int_{\mathrm{1}} ^{\mathrm{0}} \frac{{lnt}}{\mathrm{1}+{t}}{dt}\:=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnt}}{\mathrm{1}+{t}}{dt}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$
Answered by Mathspace last updated on 23/Jun/22
$$={lim}_{{x}\rightarrow\mathrm{1}} \:\:\:\frac{\int_{\mathrm{1}} ^{{x}} \frac{{lnt}}{\mathrm{1}+{t}}{dt}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$={lim}_{{x}\rightarrow\mathrm{1}} \frac{{lnx}}{\left(\mathrm{1}+{x}\right)\mathrm{2}\left({x}−\mathrm{1}\right)} \\ $$$$={lim}_{{x}\rightarrow\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}×\frac{{lnx}}{{x}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{1}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$