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Question Number 66627 by behi83417@gmail.com last updated on 17/Aug/19

Commented by behi83417@gmail.com last updated on 17/Aug/19

reposted! Q#62227

$$\mathrm{reposted}!\:\:\mathrm{Q}#\mathrm{62227} \\ $$

Commented by mathmax by abdo last updated on 18/Aug/19

1) let I =∫(√(1+x+x^2 +x^3 ))dx ⇒I =∫ (√(1+x +x^2 (1+x)))dx =∫(√((1+x)(1+x^2 )))dx=∫(√(1+x))(√(1+x^2 ))dx changement (√(1+x))=t give 1+x =t^2 ⇒x =t^2 −1 ⇒I =∫t(√(1+(t^2 −1)^2 ))(2t)dt =2 ∫ t^2 (√(1+(t^2 −1)^2 ))dt let t^2 −1 =sh(u) ⇒t^2 =1+sh(u) ⇒ 2tdt =ch du ⇒I =∫ (√(1+sh(u)))ch(u)ch(u)du =∫ ch^2 (u)(√(1+sh(u)))du =∫ ch(u)ch(u)(1+sh(u)^(1/2) du by parts f=chu and g^′ =ch(u)(1+sh(u))^(1/2) ⇒g=(2/3)(1+sh(u))^(3/2) ⇒ I =(2/3)ch(u)(1+sh(u))^(3/2) −(2/3)∫ sh(u)(1+sh(u))^(3/2) du ....be continued...

$$\left.\mathrm{1}\right)\:{let}\:{I}\:=\int\sqrt{\mathrm{1}+{x}+{x}^{\mathrm{2}} \:+{x}^{\mathrm{3}} }{dx}\:\Rightarrow{I}\:=\int\:\sqrt{\mathrm{1}+{x}\:+{x}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}{dx} \\ $$$$=\int\sqrt{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}=\int\sqrt{\mathrm{1}+{x}}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:{changement}\:\sqrt{\mathrm{1}+{x}}={t} \\ $$$${give}\:\mathrm{1}+{x}\:={t}^{\mathrm{2}} \:\Rightarrow{x}\:={t}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow{I}\:=\int{t}\sqrt{\mathrm{1}+\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{2}\:\int\:{t}^{\mathrm{2}} \sqrt{\mathrm{1}+\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\:\:\:{let}\:{t}^{\mathrm{2}} −\mathrm{1}\:={sh}\left({u}\right)\:\Rightarrow{t}^{\mathrm{2}} \:=\mathrm{1}+{sh}\left({u}\right)\:\Rightarrow \\ $$$$\mathrm{2}{tdt}\:={ch}\:{du}\:\Rightarrow{I}\:=\int\:\sqrt{\mathrm{1}+{sh}\left({u}\right)}{ch}\left({u}\right){ch}\left({u}\right){du} \\ $$$$=\int\:\:{ch}^{\mathrm{2}} \left({u}\right)\sqrt{\mathrm{1}+{sh}\left({u}\right)}{du}\:=\int\:{ch}\left({u}\right){ch}\left({u}\right)\left(\mathrm{1}+{sh}\left({u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{du}\right. \\ $$$${by}\:{parts}\:{f}={chu}\:{and}\:{g}^{'} ={ch}\left({u}\right)\left(\mathrm{1}+{sh}\left({u}\right)\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\Rightarrow{g}=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}+{sh}\left({u}\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\Rightarrow\:{I}\:=\frac{\mathrm{2}}{\mathrm{3}}{ch}\left({u}\right)\left(\mathrm{1}+{sh}\left({u}\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:−\frac{\mathrm{2}}{\mathrm{3}}\int\:\:{sh}\left({u}\right)\left(\mathrm{1}+{sh}\left({u}\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:{du} \\ $$$$....{be}\:{continued}... \\ $$

Commented by Tanmay chaudhury last updated on 18/Aug/19

challenging intregal...

$${challenging}\:{intregal}... \\ $$

Commented by Tanmay chaudhury last updated on 18/Aug/19

sir Behi pls upload the solution...if available these problems snatched our sleep..

$${sir}\:{Behi}\:{pls}\:{upload}\:{the}\:{solution}...{if}\:{available} \\ $$$${these}\:{problems}\:{snatched}\:{our}\:{sleep}.. \\ $$

Answered by mind is power last updated on 18/Aug/19

the first can′t be expressed usual function eleptic integral

$${the}\:{first}\:{can}'{t}\:{be}\:{expressed}\:{usual}\:{function}\:{eleptic}\:{integral} \\ $$$$ \\ $$

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