Question Number 125240 by hatakekakashi1729gmailcom last updated on 09/Dec/20
Commented by Dwaipayan Shikari last updated on 09/Dec/20
$$\int_{\mathrm{0}} ^{\mathrm{1}} {xlog}\left({x}\right)\:{dx}\:\:\:\:\:{logx}={t}\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{{dt}}{{dx}} \\ $$$$=\int_{−\infty} ^{\mathrm{0}} {te}^{\mathrm{2}{t}} \:\:\:{dt}\:\:\:\:\:\mathrm{2}{t}=−{u} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} {ue}^{−{u}} {du}=−\frac{\mathrm{1}}{\mathrm{4}}\Gamma\left(\mathrm{2}\right)=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$