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Question Number 83123 by M±th+et£s last updated on 28/Feb/20
∫(((x^2 −1))/(((√(x^2 +1)))(x^2 +2x−2))) dx
$$\int\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{2}\right)}\:{dx} \\ $$
Commented by mathmax by abdo last updated on 28/Feb/20
A =∫ ((x^2 −1)/( (√(x^2 +1))(x^2 +2x−2)))dx changement x=sh(t)give A =∫ ((sh^2 t−1)/(cht(sh^2 t+2sht−2)))chtdt =∫ ((((ch(2t)−1)/2)−1)/(((ch(2t)−1)/2) +2sh(t)−2))dt =∫ ((ch(2t)−3)/(ch(2t)+4sht−5))dt =∫((((e^(2t) +e^(−2t) )/2)−3)/(((e^(2t) +e^(−2t) )/2)+2(e^t −e^(−t) )−5))dt =∫ ((e^(2t) +e^(−2t) −6)/(e^(2t) +e^(−2t) +4e^t −4e^(−t) −10))dt =_(e^t =u) ∫ ((u^2 +u^(−2) −6)/(u^2 +u^(−2) +4u−4u^(−1) −10))(du/u) = ∫ ((u^4 +1 −6u^2 )/(u(u^4 +1+4u^3 −4u −10u^2 )))du =∫ ((u^4 −6u^2 +1)/(u(u^4 +4u^3 −10u^2 −4u +1)))du let decompose F(u) =((u^4 −6u^2 +1)/(u(u^4 +4u^3 −10u^2 −4u +1)))....be continued...
$${A}\:=\int\:\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{2}\right)}{dx}\:{changement}\:{x}={sh}\left({t}\right){give} \\ $$$${A}\:=\int\:\frac{{sh}^{\mathrm{2}} {t}−\mathrm{1}}{{cht}\left({sh}^{\mathrm{2}} {t}+\mathrm{2}{sht}−\mathrm{2}\right)}{chtdt} \\ $$$$=\int\:\:\frac{\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}−\mathrm{1}}{\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}\:+\mathrm{2}{sh}\left({t}\right)−\mathrm{2}}{dt} \\ $$$$=\int\:\:\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{3}}{{ch}\left(\mathrm{2}{t}\right)+\mathrm{4}{sht}−\mathrm{5}}{dt}\:\:=\int\frac{\frac{{e}^{\mathrm{2}{t}} +{e}^{−\mathrm{2}{t}} }{\mathrm{2}}−\mathrm{3}}{\frac{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}}+\mathrm{2}\left({e}^{{t}} −{e}^{−{t}} \right)−\mathrm{5}}{dt} \\ $$$$=\int\:\:\:\frac{{e}^{\mathrm{2}{t}} +{e}^{−\mathrm{2}{t}} −\mathrm{6}}{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} +\mathrm{4}{e}^{{t}} −\mathrm{4}{e}^{−{t}} −\mathrm{10}}{dt} \\ $$$$=_{{e}^{{t}} ={u}} \:\:\:\:\int\:\:\frac{{u}^{\mathrm{2}} \:+{u}^{−\mathrm{2}} −\mathrm{6}}{{u}^{\mathrm{2}} \:+{u}^{−\mathrm{2}} \:+\mathrm{4}{u}−\mathrm{4}{u}^{−\mathrm{1}} −\mathrm{10}}\frac{{du}}{{u}} \\ $$$$=\:\int\:\:\frac{{u}^{\mathrm{4}} \:+\mathrm{1}\:−\mathrm{6}{u}^{\mathrm{2}} }{{u}\left({u}^{\mathrm{4}} \:+\mathrm{1}+\mathrm{4}{u}^{\mathrm{3}} −\mathrm{4}{u}\:−\mathrm{10}{u}^{\mathrm{2}} \right)}{du} \\ $$$$=\int\:\:\frac{{u}^{\mathrm{4}} −\mathrm{6}{u}^{\mathrm{2}} \:+\mathrm{1}}{{u}\left({u}^{\mathrm{4}} \:+\mathrm{4}{u}^{\mathrm{3}} \:−\mathrm{10}{u}^{\mathrm{2}} −\mathrm{4}{u}\:+\mathrm{1}\right)}{du}\:{let}\:{decompose} \\ $$$${F}\left({u}\right)\:=\frac{{u}^{\mathrm{4}} −\mathrm{6}{u}^{\mathrm{2}} \:+\mathrm{1}}{{u}\left({u}^{\mathrm{4}} +\mathrm{4}{u}^{\mathrm{3}} −\mathrm{10}{u}^{\mathrm{2}} −\mathrm{4}{u}\:+\mathrm{1}\right)}….{be}\:{continued}… \\ $$
Commented by MJS last updated on 28/Feb/20
u^4 +4u^3 −10u^2 −4u+1= =(u−α)(u−β)(u−γ)(u−δ) α=−1−(√3)−(√(5+2(√3))) β=−1−(√3)+(√(5+2(√3))) γ=−1+(√3)−(√(5−2(√3))) δ=−1+(√3)+(√(5−2(√3))) ...
$${u}^{\mathrm{4}} +\mathrm{4}{u}^{\mathrm{3}} −\mathrm{10}{u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{1}= \\ $$$$=\left({u}−\alpha\right)\left({u}−\beta\right)\left({u}−\gamma\right)\left({u}−\delta\right) \\ $$$$\alpha=−\mathrm{1}−\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$\beta=−\mathrm{1}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$\gamma=−\mathrm{1}+\sqrt{\mathrm{3}}−\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$\delta=−\mathrm{1}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$… \\ $$
Commented by M±th+et£s last updated on 28/Feb/20
thank you for every one for the solution
$${thank}\:{you}\:{for}\:{every}\:{one}\:{for}\:{the}\:{solution} \\ $$
Commented by mathmax by abdo last updated on 28/Feb/20
thank you sir mjs
$${thank}\:{you}\:{sir}\:{mjs} \\ $$
Commented by MJS last updated on 28/Feb/20
sorry there had been a typo, now corrected
$$\mathrm{sorry}\:\mathrm{there}\:\mathrm{had}\:\mathrm{been}\:\mathrm{a}\:\mathrm{typo},\:\mathrm{now}\:\mathrm{corrected} \\ $$
Commented by MJS last updated on 28/Feb/20
it′s hard to decompose fractions like this one, not the method but all these roots...
$$\mathrm{it}'\mathrm{s}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{decompose}\:\mathrm{fractions}\:\mathrm{like}\:\mathrm{this}\:\mathrm{one}, \\ $$$$\mathrm{not}\:\mathrm{the}\:\mathrm{method}\:\mathrm{but}\:\mathrm{all}\:\mathrm{these}\:\mathrm{roots}… \\ $$

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