Question Number 66627 by behi83417@gmail.com last updated on 17/Aug/19
Commented by behi83417@gmail.com last updated on 17/Aug/19
$$\mathrm{reposted}!\:\:\mathrm{Q}#\mathrm{62227} \\ $$
Commented by mathmax by abdo last updated on 18/Aug/19
$$\left.\mathrm{1}\right)\:{let}\:{I}\:=\int\sqrt{\mathrm{1}+{x}+{x}^{\mathrm{2}} \:+{x}^{\mathrm{3}} }{dx}\:\Rightarrow{I}\:=\int\:\sqrt{\mathrm{1}+{x}\:+{x}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}{dx} \\ $$$$=\int\sqrt{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}=\int\sqrt{\mathrm{1}+{x}}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:{changement}\:\sqrt{\mathrm{1}+{x}}={t} \\ $$$${give}\:\mathrm{1}+{x}\:={t}^{\mathrm{2}} \:\Rightarrow{x}\:={t}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow{I}\:=\int{t}\sqrt{\mathrm{1}+\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{2}\:\int\:{t}^{\mathrm{2}} \sqrt{\mathrm{1}+\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\:\:\:{let}\:{t}^{\mathrm{2}} −\mathrm{1}\:={sh}\left({u}\right)\:\Rightarrow{t}^{\mathrm{2}} \:=\mathrm{1}+{sh}\left({u}\right)\:\Rightarrow \\ $$$$\mathrm{2}{tdt}\:={ch}\:{du}\:\Rightarrow{I}\:=\int\:\sqrt{\mathrm{1}+{sh}\left({u}\right)}{ch}\left({u}\right){ch}\left({u}\right){du} \\ $$$$=\int\:\:{ch}^{\mathrm{2}} \left({u}\right)\sqrt{\mathrm{1}+{sh}\left({u}\right)}{du}\:=\int\:{ch}\left({u}\right){ch}\left({u}\right)\left(\mathrm{1}+{sh}\left({u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{du}\right. \\ $$$${by}\:{parts}\:{f}={chu}\:{and}\:{g}^{'} ={ch}\left({u}\right)\left(\mathrm{1}+{sh}\left({u}\right)\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\Rightarrow{g}=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}+{sh}\left({u}\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\Rightarrow\:{I}\:=\frac{\mathrm{2}}{\mathrm{3}}{ch}\left({u}\right)\left(\mathrm{1}+{sh}\left({u}\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:−\frac{\mathrm{2}}{\mathrm{3}}\int\:\:{sh}\left({u}\right)\left(\mathrm{1}+{sh}\left({u}\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:{du} \\ $$$$….{be}\:{continued}… \\ $$
Commented by Tanmay chaudhury last updated on 18/Aug/19
$${challenging}\:{intregal}… \\ $$
Commented by Tanmay chaudhury last updated on 18/Aug/19
$${sir}\:{Behi}\:{pls}\:{upload}\:{the}\:{solution}…{if}\:{available} \\ $$$${these}\:{problems}\:{snatched}\:{our}\:{sleep}.. \\ $$
Answered by mind is power last updated on 18/Aug/19
$${the}\:{first}\:{can}'{t}\:{be}\:{expressed}\:{usual}\:{function}\:{eleptic}\:{integral} \\ $$$$ \\ $$