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




Question Number 38720 by maxmathsup by imad last updated on 28/Jun/18
find   ∫     (((√(x+1)) −(√(x−1)))/( (√(x+1)) −(√(x−1))))dx
findx+1x1x+1x1dx
Commented by math khazana by abdo last updated on 28/Jun/18
the Q is find  ∫    (((√(x+1)) −(√(x−1)))/( (√(x+1)) +(√(x−1))))dx
theQisfindx+1x1x+1+x1dx
Commented by prof Abdo imad last updated on 29/Jun/18
I= ∫   (({(√(x+1)) −(√(x−1))}^2 )/2)dx  = ∫    ((x+1 −2(√(x^2  −1))+x−1)/2)dx  =∫   xdx −∫   (√(x^2 −1))dx  =(x^2 /2) − ∫ (√(x^2 −1))dx but changement x=ch(t)  give ∫(√(x^2 −1))dx=∫  sht shtdt  =∫ sh^2 t dt =∫  ((ch(2t)−1)/2)dt  =(1/4)sh(2t) −(t/2) =(1/4)2sh(t)ch(t) −(t/2)  =(1/2)x(√(x^2 −1))   −(1/2)argch(x)  =(x/2)(√(x^2 −1))  −(1/2)ln(x +(√(x^2 −1))) ⇒  I = (x^2 /2)  −(x/2)(√(x^2 −1))  +(1/2)ln(x+(√(x^2 −1))) +c
I={x+1x1}22dx=x+12x21+x12dx=xdxx21dx=x22x21dxbutchangementx=ch(t)givex21dx=shtshtdt=sh2tdt=ch(2t)12dt=14sh(2t)t2=142sh(t)ch(t)t2=12xx2112argch(x)=x2x2112ln(x+x21)I=x22x2x21+12ln(x+x21)+c

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