## anonymous 4 years ago integrate (x+1)/(x-1)

1. anonymous

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2. anonymous

${x+1 \over x-1} = {(x-1) + 2 \over x-1} = 1 + {2 \over x-1}$ 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.

3. anonymous