Question Number 152828 by mathdanisur last updated on 01/Sep/21
$$\underset{\:\mathrm{1}} {\overset{\:\infty} {\int}}\:\frac{\sqrt{\mathrm{x}}\:\mathrm{ln}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:\mathrm{dx}\:=\:? \\ $$
Answered by Ar Brandon last updated on 01/Sep/21
$${f}\left(\kappa\right)=\int_{\mathrm{1}} ^{\infty} \frac{{x}^{\kappa} }{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}=\int_{\mathrm{1}} ^{\infty} \frac{{x}^{\kappa} \left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}−{x}^{\mathrm{3}} \right)}{dx} \\ $$$${x}={u}^{−\frac{\mathrm{1}}{\mathrm{3}}} \Rightarrow{dx}=−\frac{\mathrm{1}}{\mathrm{3}}{u}^{−\frac{\mathrm{4}}{\mathrm{3}}} {du} \\ $$$${f}\left(\kappa\right)=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{−\frac{\kappa}{\mathrm{3}}} −{u}^{−\frac{\kappa+\mathrm{1}}{\mathrm{3}}} }{\mathrm{1}−{u}^{−\mathrm{1}} }\centerdot{u}^{−\frac{\mathrm{4}}{\mathrm{3}}} {du} \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{−\frac{\kappa+\mathrm{1}}{\mathrm{3}}} −{u}^{−\frac{\kappa+\mathrm{2}}{\mathrm{3}}} }{\mathrm{1}−{u}}{du} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}}\left(\Phi\left(\frac{\mathrm{2}−\kappa}{\mathrm{3}}\right)−\Phi\left(\frac{\mathrm{1}−\kappa}{\mathrm{3}}\right)\right. \\ $$$${f}\:'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{9}}\left(\Phi'\left(\frac{\mathrm{1}}{\mathrm{6}}\right)−\Phi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$
Commented by mathdanisur last updated on 02/Sep/21
$$\mathrm{Thank}\:\mathrm{You}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$