I need intergral for this (2x-4)*(x-1)^{-2 } the answer is (2x-2)- 2 ln (x-1) + c
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$$\int \frac{2x-4}{(x-1)^2} dx = 2 \int \frac{x-2}{(x-1)^2}dx= 2 \int \frac{x-1 -1 }{(x-1)^2}dx = 2 \int \left(\frac{1}{x-1} - \frac{1}{(x-1)^2}\right)dx$$ Now let $u= x-1$, $du= dx$, can you do the rest?