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mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1337384917720:dw

lolampig
 2 years ago
Best 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.

Silenthill
 2 years ago
Best ResponseYou've already chosen the best response.0very helpful thank you
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