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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: Most challenging one!!!
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[Post New] 24 Feb 2007 13:11:48 IST
   Subject: Re:Most challenging one!!!
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CyBorG (1574)

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Olaaa!! Perrrfect answer. 252  [409 rates]
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Check my method Shakir
Put 1+x4=x4y4
x4=1/(y4-1)
dx= -y3dy/x3(y4-1)2
So given integral is  -y3dy/[x4y(y4-1)2]= -y2dy/x4(y4-1)2  
= -y2dy/(y4-1)
So final answer is (1/2)[(1/2)log{(1+y)/(1-y)]-tan-1y}+c
So which one you think is shorter???
 
 
 
-ADARSH NITK Surathkal
  this reply:   9 points  (with 1  Olaaa!! Perrrfect answer.   in 3   votes   )     [?]
 
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