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

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Question Number 53294 by gunawan last updated on 20/Jan/19
∫_(−1/2) ^(1/2) [(((x+1)/(x−1)))^2 +(((x−1)/(x+1)))^2 −2]^(1/2) dx=...
∫−1/21/2[(x+1x−1)2+(x−1x+1)2−2]1/2dx=…
Answered by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19
f(x)=[(((x+1)/(x−1)))^2 +(((x−1)/(x+1)))^2 −2]^(1/2) =[{(((x+1)/(x−1)))−(((x−1)/(x+1)))}^2 ]^(1/2) =(((x+1)^2 −(x−1)^2 )/((x^2 −1))) =((4x)/(x^2 −1)) ∫_((−1)/2) ^(1/2) ((4x)/(x^2 −1))dx←look here f(−x)=((−4x)/(x^2 −1))=−f(x) so ∫_((−1)/2) ^(1/2) ((4x)/(x^2 −1))=0 if we clculate.. 2∫_((−1)/2) ^(1/2) ((d(x^2 −1))/(x^2 −1)) 2×∣ln(x^2 −1)∣_((−1)/2) ^(1/2) 2[ln∣((1/4)−1)∣−ln∣((1/4)−1)∣]=0
f(x)=[(x+1x−1)2+(x−1x+1)2−2]12=[{(x+1x−1)−(x−1x+1)}2]12=(x+1)2−(x−1)2(x2−1)=4xx2−1∫−12124xx2−1dx←lookheref(−x)=−4xx2−1=−f(x)so∫−12124xx2−1=0ifweclculate..2∫−1212d(x2−1)x2−12×∣ln(x2−1)∣−12122[ln∣(14−1)∣−ln∣(14−1)∣]=0

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