Question Number 138132 by mathmax by abdo last updated on 10/Apr/21
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{logx}}{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{2}}\mathrm{dx} \\ $$
Answered by Ñï= last updated on 13/Apr/21
$${I}=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\:{x}}{{x}^{\mathrm{2}} −{x}+\mathrm{2}}{dx}=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{ln}\:\left(\sqrt{\mathrm{2}}{x}\right)}{\mathrm{2}{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{2}}{dx}=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}+{lnx}}{\mathrm{2}{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{2}}{dx} \\ $$$$=\frac{{ln}\mathrm{2}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{2}{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{2}}+\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{\mathrm{2}{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{2}}{dx}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{−{lnx}}{\mathrm{2}−\sqrt{\mathrm{2}}{x}+\mathrm{2}{x}^{\mathrm{2}} }{dx}=−{I}_{\mathrm{2}} \\ $$$$\Rightarrow{I}_{\mathrm{2}} =\mathrm{0} \\ $$$${I}={I}_{\mathrm{1}} =\frac{{ln}\mathrm{2}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\sqrt{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{4}}}=\frac{{ln}\mathrm{2}}{\:\mathrm{2}\sqrt{\frac{\mathrm{7}}{\mathrm{4}}}}\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{2}}}{\:\sqrt{\frac{\mathrm{7}}{\mathrm{4}}}}\mid_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\left(\frac{\pi}{\mathrm{2}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\right){ln}\:\mathrm{2} \\ $$