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find-x-2-x-2-1-dx-




Question Number 36424 by abdo.msup.com last updated on 01/Jun/18
find ∫  x^2 (√(x^2 −1))dx
findx2x21dx
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=14sh2(2t)dt=18(ch(4t)1)dt=t8+132sh(4t)=18argch(x)+132sh{4argch(x)}=18ln(x+x21)+132sh{4ln(x+x21)}butsh{4ln(x+x21)=sh{ln(x+x21)2}=(x+x21)2+(x+x21)22I=164{(x+x21)2+(x+x21)2}18ln(x+x21).
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+x21)2(x+x21)2}18ln(x+x21).

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