Question Number 38746 by MJS last updated on 29/Jun/18
$$\mathrm{this}\:\mathrm{is}\:\mathrm{still}\:\mathrm{waiting}\:\mathrm{to}\:\mathrm{be}\:\mathrm{solved}… \\ $$$$\int\frac{\sqrt{\left({t}−\mathrm{1}\right){t}\left({t}+\mathrm{1}\right)}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt}=? \\ $$
Commented by behi83417@gmail.com last updated on 29/Jun/18
$${this}\:{can}\:{not}\:{be}\:{difined}\:\:{in}\:{terms}\:{of} \\ $$$${primary}\:{functions}.{dont}\:{spend}\:{time}\:{on} \\ $$$${it}. \\ $$
Commented by prof Abdo imad last updated on 29/Jun/18
$${I}\:=\:\int\:\:\:\frac{\sqrt{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt}\:\:{changement}\:{t}={ch}\left({x}\right){give} \\ $$$${I}\:\:=\:\int\:\:\frac{\sqrt{{ch}\left({x}\right)}\:{sh}\left({x}\right)}{\mathrm{3}{ch}^{\mathrm{2}} {x}\:−\mathrm{4}}\:{sh}\left({x}\right){dx} \\ $$$$=\:\int\:\:\frac{{sh}^{\mathrm{2}} \left({x}\right)\sqrt{{ch}\left({x}\right)}}{\mathrm{3}{ch}^{\mathrm{2}} {x}\:−\mathrm{4}}{dx}\:\:\sqrt{{chx}}={u}\:\Rightarrow{chx}={u}^{\mathrm{2}} \Rightarrow{x}={argch}\left({u}^{\mathrm{2}} \right) \\ $$$${I}\:=\:\int\:\:\frac{\left({u}^{\mathrm{4}} −\mathrm{1}\right){u}}{\mathrm{3}{u}^{\mathrm{4}} −\mathrm{4}}\:\:\frac{\mathrm{2}{udu}}{\:\sqrt{{u}^{\mathrm{4}} −\mathrm{1}}} \\ $$$$=\:\int\:\:\:\frac{\mathrm{2}{u}^{\mathrm{2}} \sqrt{{u}^{\mathrm{4}} −\mathrm{1}}}{\mathrm{3}{u}^{\mathrm{4}} −\mathrm{4}}{du}=\:\int\:\:\frac{\mathrm{2}{u}^{\mathrm{2}} \sqrt{{u}^{\mathrm{4}} −\mathrm{1}}}{\mathrm{4}{u}^{\mathrm{4}} −\mathrm{4}\:−{u}^{\mathrm{4}} }{du} \\ $$$$=\int\:\:\:\:\frac{\mathrm{2}{u}^{\mathrm{2}} \sqrt{{u}^{\mathrm{4}} −\mathrm{1}}}{\mathrm{4}\left(\sqrt{{u}^{\mathrm{4}} −\mathrm{1}}\right)^{\mathrm{2}} \:−{u}^{\mathrm{4}} }{du}\:\:{chang}.\sqrt{{u}^{\mathrm{4}} −\mathrm{1}}={x}\Rightarrow{u}^{\mathrm{4}} −\mathrm{1}={x}^{\mathrm{2}} \Rightarrow{u}=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${I}\:=\:\int\:\:\:\frac{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{4}{x}^{\mathrm{2}} \:−\mathrm{1}−{x}^{\mathrm{2}} }\:\frac{{x}}{\mathrm{2}}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{4}}} \:{dx} \\ $$$$=\:\int\:\:\:\frac{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)^{−\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}}{dx}\:{chang}\:.{x}={tan}\theta\:{give} \\ $$$${I}\:=\:\int\:\:\:\frac{{tan}^{\mathrm{2}} \theta\:\left({cos}^{−\mathrm{2}} \theta\right)^{−\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{3}{tan}^{\mathrm{2}} \theta\:−\mathrm{1}}\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$….{be}\:{continued}… \\ $$
Commented by MJS last updated on 29/Jun/18
$$\mathrm{Sir}\:\mathrm{Behi},\:\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{same}.\:\mathrm{but}\:\mathrm{can}\:\mathrm{we}\:\mathrm{be}\:\mathrm{sure}? \\ $$