Question Number 209521 by mokys last updated on 12/Jul/24
$${find}\:{the}\:{integral}\:\int\:\frac{{dx}}{{x}^{\mathrm{4}} +{a}^{\mathrm{4}} }\:{by}\:{complex}\:{number}\:?\: \\ $$
Answered by mathmax last updated on 14/Jul/24
$${I}=\int\:\frac{{dx}}{{x}^{\mathrm{4}} −{i}^{\mathrm{2}} {a}^{\mathrm{4}} }=\int\frac{{dx}}{\left({x}^{\mathrm{2}} −{ia}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{ia}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ia}^{\mathrm{2}} }\int\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{ia}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{ia}^{\mathrm{2}} }\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ia}^{\mathrm{2}} }\left\{\int\frac{{dx}}{{x}^{\mathrm{2}} −{ia}^{\mathrm{2}} }−\int\frac{{dx}}{{x}^{\mathrm{2}} +{ia}^{\mathrm{2}} }\right\}{dx} \\ $$$${I}_{\mathrm{1}} =\int\:\:\frac{{dx}}{\left({x}−\sqrt{{i}}{a}\right)\left({x}+\sqrt{{i}}{a}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{i}}{a}}\int\left(\frac{\mathrm{1}}{{x}−\sqrt{{i}}{a}}−\frac{\mathrm{1}}{\left.{x}+\sqrt{{i}}{a}\right)}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{i}}{a}}{ln}\left(\frac{{x}−\sqrt{{i}}{a}}{{x}+\sqrt{{i}}{a}}\right)\:+{c}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\int\:\frac{{dx}}{{x}^{\mathrm{2}} +{ia}^{\mathrm{2}} }=\int\frac{{dx}}{{x}^{\mathrm{2}} −\left(−{i}\right){a}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{i}}{a}}\int\left(\frac{\mathrm{1}}{{x}−\sqrt{−{i}}{a}}−\frac{\mathrm{1}}{{x}+\sqrt{−{i}}{a}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{i}}{a}}{ln}\left(\frac{{x}−\sqrt{−{i}}{a}}{{x}+\sqrt{−{i}}{a}}\right)+{c}_{\mathrm{2}} \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{2}{ia}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{2}\sqrt{{i}}{a}}{ln}\left(\frac{{x}−\sqrt{{i}}{a}}{{x}+\sqrt{{i}}{a}}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}{ia}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{i}}{a}}{ln}\left(\frac{{x}−\sqrt{−{i}}{a}}{{x}+\sqrt{−{i}}{a}}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{a}^{\mathrm{3}} {i}\sqrt{{i}}}{ln}\left(\frac{{x}−\sqrt{{i}}{a}}{{x}+\sqrt{{i}}{a}}\right)−\frac{\mathrm{1}}{\mathrm{4}{a}^{\mathrm{3}} {i}\sqrt{−{i}}}{ln}\left(\frac{{x}−\sqrt{−{i}}{a}}{{x}+\sqrt{−{i}}{a}}\right)+{C} \\ $$