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[Post New] 7 Apr 2010 00:30:03 IST
   Subject: ∫(e1/x+1)/x2 dx
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Himsy (22)

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∫(e1/x+1)/x2 dx
    
[Post New] 7 Apr 2010 17:40:08 IST
   Subject: ∫(e1/x+1)/x2 dx
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Deepak Aggarwal (3759)

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I = ∫(e1/x+1) / x2  dx

Put y = 1/x

So, - dy = dx / x2

So, I = - ∫ (ey + 1)dy

I = - (ey + y) + C

I = - e1/x - (1/x) + C

 
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[Post New] 7 Apr 2010 17:42:04 IST
   Subject: ∫(e1/x+1)/x2 dx
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Debmalya Ray (66)

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let, 1/x=z

 

or,1/x^2 dx=-dz

 

integrate -(e^z+1)dz

 

=-e^z-z+c (c is an arbitary constant)

 

=-e^1/x-1/x+c

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[Post New] 8 Apr 2010 00:40:09 IST
   Subject: ∫(e1/x+1)/x2 dx
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illuminatti (14)

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Possible intermediate steps:\n integral (1+e^(1/x))/x^2 dx\nFor the integrand (1+e^(1/x))/x^2, substitute u = 1/x and du = -1/x^2 dx:\n = - integral (e^u+1) du\nIntegrate the sum term by term:\n = - integral 1 du- integral e^u du\nThe integral of e^u is e^u:\n = -e^u- integral 1 du\nThe integral of 1 is u:\n = -u-e^u+constant\nSubstitute back for u = 1/x:\n = -(e^(1/x) x+1)/x+constant\nWhich is equal to:\n = -e^(1/x)-1/x+constant
illuminatti different uncountable state
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