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Silenthill
integrate (x+1)/(x-1)
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\[{x+1 \over x-1} = {(x-1) + 2 \over x-1} = 1 + {2 \over x-1}\] Now we let u = x-1, so that du/dx = 1, and du = dx. Then \[\int\limits 1 + {2 \over x-1} dx = \int\limits 1 + {2 \over u} du\]The integral of 1/u is ln|u|, and the integral of 1 is u, so we have the integral to be \[u + 2\ln|u| + c = (x-1) + 2\ln|x-1| + c =x + 2\ln|x-1| + c\]Where the -1 was absorbed into the constant of integration in the last step.
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