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Integral Calculus

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 Joined: 3 Oct 2011 Post: 82
8 May 2012 18:48:26 IST
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integration
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Integral Calculus

log(1+x)/1+x

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Joined: 8 Oct 2011
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8 May 2012 19:28:21 IST
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(log(1+x))^2/2

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Joined: 3 May 2012
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8 May 2012 22:36:57 IST
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out u=1+x then use parts method,,,,, first take logarithemic function

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8 May 2012 22:46:15 IST
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put log(1+x)=t then dx/(1+x)=dt so u have integration of tdt=t^2/2=[log(1+x)]^2/2

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9 May 2012 09:40:53 IST
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Right answer by mihir and right solution by tarun. By substituting 1+x = t you will get [{log(1+x)}^2]/2 as answer.

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9 May 2012 09:44:06 IST
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gud work by tarun............rit ans is (log(1+x))^2/2 simple substitutn

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