Question Number 141244 by ajfour last updated on 17/May/21
$${I}=\int_{\mathrm{0}} ^{\:\infty} \frac{{x}\left\{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){x}−\mathrm{2}{a}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} \right\}{dx}}{\left({a}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} \left\{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){x}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right\}} \\ $$
Commented by ajfour last updated on 17/May/21
this will get me perimeter of ellipse!
Answered by MJS_new last updated on 17/May/21
$${I}_{\mathrm{1}} =\int\frac{{x}}{\left({a}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=−\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} \\ $$$${I}_{\mathrm{2}} =−\frac{\mathrm{3}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\int\frac{{dx}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{2}} }=−\frac{\mathrm{3arctan}\:\frac{{ax}}{{b}}}{{a}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){b}} \\ $$$$ \\ $$$${I}_{\mathrm{3}} =\frac{\mathrm{3}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\int\frac{{dx}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){x}+{b}^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{3}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\sqrt{{a}^{\mathrm{4}} −\mathrm{6}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{4}} }}\mathrm{ln}\:\frac{\mathrm{2}{a}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\sqrt{{a}^{\mathrm{4}} −\mathrm{6}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{4}} }}{\mathrm{2}{a}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\sqrt{{a}^{\mathrm{4}} −\mathrm{6}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{4}} }} \\ $$$$\mathrm{now}\:\mathrm{calculate}\:\mathrm{within}\:\mathrm{the}\:\mathrm{borders} \\ $$