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

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