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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∫x−2x2−1x+2x2−1dx
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 .....
letI=∫x−2x2−1x+2x2−1dxchangementx=ch(t)giveI=∫ch(t)−2sh(t)ch(t)+2sh(t)sh(t)dt=∫ch(t)sh(t)−2sh2(t)ch(t)+2sh(t)dt=∫12sh(2t)−2ch(2t)−12ch(t)+2sh(t)dt=12∫sh(2t)−2ch(2t)+2ch(t)+2sh(t)dt2I=∫e2t−e−2t2−2e2t+e−2t2+2et+e−t2+2et−e−t2dt=∫e2t−e−2t−2e2t−2e−2t+4et+e−t+2et−2e−tdt=∫=4−e2t−3e−2t3et−e−tdt=et=z∫4−z2−3z−23z−z−1dzz=∫4−z2−3z−23z2−1dz=∫4z2−z4−33z4−z2dzletdecomposeF(z)=4z2−z4−3z2(3z2−1)⇒F(z)=−z4−4z2+33z4−z2=−133z4−12z2+93z4−z2=−13×3z4−z2−10z2+93z4−z2=−13−13−10z2+9z2(3z2−1)=−13+az+bz2+c3z−1+d3z+1⇒∫F(z)dz=−z3+aln∣z∣−bz+c3ln∣3z−1∣+d3ln∣3z+1∣+C=−et3+at−be−t+c3ln∣3et−1∣+d3ln∣3et+1∣+Cbutt=argch(x)=ln(x+x2−1)⇒2I=−13(x+x2−1)−bx+x2−1+c3ln∣3(x+x2−1)−1∣+d3ln∣3(x+x2−1)+1∣+Cresttocalculatethecoeff.ci…..

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