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Question Number 67005 by mathmax by abdo last updated on 21/Aug/19
find ∫ ((x−2(√(x^2 −1)))/(x+2(√(x^2 −1))))dx
$${find}\:\int\:\:\:\frac{{x}−\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dx} \\ $$
Commented by mathmax by abdo last updated on 27/Aug/19
let I =∫ ((x−2(√(x^2 −1)))/(x+2(√(x^2 −1))))dx changement x =ch(t)give I =∫ ((ch(t)−2sh(t))/(ch(t)+2sh(t))) sh(t)dt =∫ ((ch(t)sh(t)−2sh^2 (t))/(ch(t)+2sh(t)))dt =∫ (((1/2)sh(2t)−2 ((ch(2t)−1)/2))/(ch(t)+2sh(t)))dt =(1/2) ∫ ((sh(2t)−2ch(2t)+2)/(ch(t)+2sh(t)))dt 2I =∫ ((((e^(2t) −e^(−2t) )/2)−2 ((e^(2t) +e^(−2t) )/2) +2)/(((e^t +e^(−t) )/2)+2 ((e^t −e^(−t) )/2))) dt = ∫ ((e^(2t) −e^(−2t) −2e^(2t) −2e^(−2t) +4)/(e^t +e^(−t) +2 e^t −2e^(−t) ))dt =∫= ((4−e^(2t) −3e^(−2t) )/(3e^t −e^(−t) )) dt =_(e^t =z) ∫ ((4−z^2 −3z^(−2) )/(3z−z^(−1) ))(dz/z) =∫ ((4−z^2 −3z^(−2) )/(3z^2 −1))dz =∫ ((4z^2 −z^4 −3)/(3z^4 −z^2 ))dz let decompose F(z) =((4z^2 −z^4 −3)/(z^2 (3z^2 −1))) ⇒ F(z) =−((z^4 −4z^2 +3)/(3z^4 −z^2 )) =−(1/3)((3z^4 −12z^2 +9)/(3z^4 −z^2 )) =−(1/3)×((3z^4 −z^2 −10z^2 +9)/(3z^4 −z^2 )) =−(1/3)−(1/3)((−10z^2 +9)/(z^2 (3z^2 −1))) =−(1/3) +(a/z) +(b/z^2 ) +(c/( (√3)z−1)) +(d/( (√3)z +1)) ⇒ ∫ F(z)dz =−(z/3) +aln∣z∣−(b/z) +(c/( (√3)))ln∣(√3)z −1∣ +(d/( (√3)))ln∣(√3)z+1∣ +C =−(e^t /3) +at −be^(−t) +(c/( (√3)))ln∣(√3)e^t −1∣ +(d/( (√3)))ln∣(√3)e^t +1∣ +C but t =argch(x)=ln(x+(√(x^2 −1))) ⇒ 2I =−(1/3)(x+(√(x^2 −1))) −(b/(x+(√(x^2 −1)))) +(c/( (√3)))ln∣(√3)(x+(√(x^2 −1)))−1∣ +(d/( (√3)))ln∣(√3)(x+(√(x^2 −1))) +1∣ +C rest to calculate the coeff.c_i .....
$${let}\:{I}\:=\int\:\:\frac{{x}−\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{dx}\:\:{changement}\:{x}\:={ch}\left({t}\right){give} \\ $$$${I}\:=\int\:\:\:\frac{{ch}\left({t}\right)−\mathrm{2}{sh}\left({t}\right)}{{ch}\left({t}\right)+\mathrm{2}{sh}\left({t}\right)}\:{sh}\left({t}\right){dt}\:=\int\:\:\:\frac{{ch}\left({t}\right){sh}\left({t}\right)−\mathrm{2}{sh}^{\mathrm{2}} \left({t}\right)}{{ch}\left({t}\right)+\mathrm{2}{sh}\left({t}\right)}{dt} \\ $$$$=\int\:\:\:\:\frac{\frac{\mathrm{1}}{\mathrm{2}}{sh}\left(\mathrm{2}{t}\right)−\mathrm{2}\:\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}}{{ch}\left({t}\right)+\mathrm{2}{sh}\left({t}\right)}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{{sh}\left(\mathrm{2}{t}\right)−\mathrm{2}{ch}\left(\mathrm{2}{t}\right)+\mathrm{2}}{{ch}\left({t}\right)+\mathrm{2}{sh}\left({t}\right)}{dt} \\ $$$$\mathrm{2}{I}\:=\int\:\:\:\frac{\frac{{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} }{\mathrm{2}}−\mathrm{2}\:\frac{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}}\:+\mathrm{2}}{\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}+\mathrm{2}\:\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}}\:{dt} \\ $$$$=\:\int\:\:\frac{{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} −\mathrm{2}{e}^{\mathrm{2}{t}} −\mathrm{2}{e}^{−\mathrm{2}{t}} +\mathrm{4}}{{e}^{{t}} \:+{e}^{−{t}} \:+\mathrm{2}\:{e}^{{t}} −\mathrm{2}{e}^{−{t}} }{dt}\:=\int=\:\:\:\frac{\mathrm{4}−{e}^{\mathrm{2}{t}} −\mathrm{3}{e}^{−\mathrm{2}{t}} }{\mathrm{3}{e}^{{t}} \:−{e}^{−{t}} }\:{dt} \\ $$$$=_{{e}^{{t}} ={z}} \:\:\:\int\:\:\:\frac{\mathrm{4}−{z}^{\mathrm{2}} −\mathrm{3}{z}^{−\mathrm{2}} }{\mathrm{3}{z}−{z}^{−\mathrm{1}} }\frac{{dz}}{{z}}\:=\int\:\:\frac{\mathrm{4}−{z}^{\mathrm{2}} −\mathrm{3}{z}^{−\mathrm{2}} }{\mathrm{3}{z}^{\mathrm{2}} −\mathrm{1}}{dz} \\ $$$$=\int\:\:\:\:\frac{\mathrm{4}{z}^{\mathrm{2}} −{z}^{\mathrm{4}} −\mathrm{3}}{\mathrm{3}{z}^{\mathrm{4}} −{z}^{\mathrm{2}} }{dz}\:\:{let}\:{decompose}\:{F}\left({z}\right)\:=\frac{\mathrm{4}{z}^{\mathrm{2}} −{z}^{\mathrm{4}} −\mathrm{3}}{{z}^{\mathrm{2}} \left(\mathrm{3}{z}^{\mathrm{2}} −\mathrm{1}\right)}\:\Rightarrow \\ $$$${F}\left({z}\right)\:=−\frac{{z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{2}} \:+\mathrm{3}}{\mathrm{3}{z}^{\mathrm{4}} −{z}^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{3}}\frac{\mathrm{3}{z}^{\mathrm{4}} −\mathrm{12}{z}^{\mathrm{2}} \:+\mathrm{9}}{\mathrm{3}{z}^{\mathrm{4}} −{z}^{\mathrm{2}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{3}{z}^{\mathrm{4}} −{z}^{\mathrm{2}} −\mathrm{10}{z}^{\mathrm{2}} \:+\mathrm{9}}{\mathrm{3}{z}^{\mathrm{4}} −{z}^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\frac{−\mathrm{10}{z}^{\mathrm{2}} \:+\mathrm{9}}{{z}^{\mathrm{2}} \left(\mathrm{3}{z}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{{a}}{{z}}\:+\frac{{b}}{{z}^{\mathrm{2}} }\:+\frac{{c}}{\:\sqrt{\mathrm{3}}{z}−\mathrm{1}}\:+\frac{{d}}{\:\sqrt{\mathrm{3}}{z}\:+\mathrm{1}}\:\:\:\Rightarrow \\ $$$$\int\:{F}\left({z}\right){dz}\:=−\frac{{z}}{\mathrm{3}}\:+{aln}\mid{z}\mid−\frac{{b}}{{z}}\:+\frac{{c}}{\:\sqrt{\mathrm{3}}}{ln}\mid\sqrt{\mathrm{3}}{z}\:−\mathrm{1}\mid\:+\frac{{d}}{\:\sqrt{\mathrm{3}}}{ln}\mid\sqrt{\mathrm{3}}{z}+\mathrm{1}\mid\:+{C} \\ $$$$=−\frac{{e}^{{t}} }{\mathrm{3}}\:+{at}\:−{be}^{−{t}} \:+\frac{{c}}{\:\sqrt{\mathrm{3}}}{ln}\mid\sqrt{\mathrm{3}}{e}^{{t}} −\mathrm{1}\mid\:+\frac{{d}}{\:\sqrt{\mathrm{3}}}{ln}\mid\sqrt{\mathrm{3}}{e}^{{t}} \:+\mathrm{1}\mid\:+{C} \\ $$$${but}\:{t}\:={argch}\left({x}\right)={ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=−\frac{\mathrm{1}}{\mathrm{3}}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:−\frac{{b}}{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:+\frac{{c}}{\:\sqrt{\mathrm{3}}}{ln}\mid\sqrt{\mathrm{3}}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)−\mathrm{1}\mid \\ $$$$+\frac{{d}}{\:\sqrt{\mathrm{3}}}{ln}\mid\sqrt{\mathrm{3}}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:+\mathrm{1}\mid\:+{C}\: \\ $$$${rest}\:{to}\:{calculate}\:{the}\:{coeff}.{c}_{{i}} ….. \\ $$

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