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mahmit2012Best ResponseYou've already chosen the best response.0
dw:1337384917720:dw
 one year ago

lolampigBest ResponseYou've already chosen the best response.6
\[{x+1 \over x1} = {(x1) + 2 \over x1} = 1 + {2 \over x1}\] Now we let u = x1, so that du/dx = 1, and du = dx. Then \[\int\limits 1 + {2 \over x1} dx = \int\limits 1 + {2 \over u} du\]The integral of 1/u is lnu, and the integral of 1 is u, so we have the integral to be \[u + 2\lnu + c = (x1) + 2\lnx1 + c =x + 2\lnx1 + c\]Where the 1 was absorbed into the constant of integration in the last step.
 one year ago

SilenthillBest ResponseYou've already chosen the best response.0
very helpful thank you
 one year ago
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