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Question Number 36424 by abdo.msup.com last updated on 01/Jun/18
find ∫ x^2 (√(x^2 −1))dx
find∫x2x2−1dx
Commented by prof Abdo imad last updated on 03/Jun/18
changement x = ch(t) give I = ∫ ch^2 t sh(t)sh(t)dt=∫ (ch(t)sh(t))^2 dt =(1/4) ∫ sh^2 (2t)dt =(1/8) ∫ (ch(4t)−1)dt = −(t/8) + (1/(32)) sh(4t) =−(1/8) argch(x) +(1/(32))sh{4argch(x)} =−(1/8)ln(x +(√(x^2 −1))) +(1/(32))sh{4ln(x +(√(x^2 −1)))} but sh{4ln(x+(√(x^2 −1)))=sh{ln(x+(√(x^2 −1)))^2 } = (((x+(√(x^2 −1)))^2 +(x+(√(x^2 −1)))^(−2) )/2) I = (1/(64)){ (x+(√(x^2 −1)))^2 +(x+(√(x^2 −1)))^(−2) } −(1/8)ln(x +(√(x^2 −1))) .
changementx=ch(t)giveI=∫ch2tsh(t)sh(t)dt=∫(ch(t)sh(t))2dt=14∫sh2(2t)dt=18∫(ch(4t)−1)dt=−t8+132sh(4t)=−18argch(x)+132sh{4argch(x)}=−18ln(x+x2−1)+132sh{4ln(x+x2−1)}butsh{4ln(x+x2−1)=sh{ln(x+x2−1)2}=(x+x2−1)2+(x+x2−1)−22I=164{(x+x2−1)2+(x+x2−1)−2}−18ln(x+x2−1).
Commented by abdo mathsup last updated on 04/Jun/18
I = (1/(64)){ (x +(√(x^2 −1)))^2 −(x +(√(x^2 −1)))^(−2) } −(1/8)ln(x +(√(x^2 −1))) .
I=164{(x+x2−1)2−(x+x2−1)−2}−18ln(x+x2−1).

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