What is the integral of:`x/(x-1)^2`

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`int x/(x-1)^2 dx `

`u= x-1 ==gt x= u+1`

du = dx

`==gt int x/(x-1)^2 dx = int (u+1)/u^2 du`

`= int u/u^2 du+ int 1/u^2 du `

`= int 1/u du + int 1/u^2 du `

`= ln u - 1/u + C = ln (x-1) - 1/(x-1) + C `

`==gt int x/(x-1)^2 dx = ln (x-1) - 1/(x-1) + C`

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