What is integral of y = 1/(x-1)?

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You need to evaluate the indefinite integral of the given function `f(x) = y = 1/(x-1)` such that:

`int f(x)dx = int (dx)/(x - 1)`

You should come up with the following substitution, such that:

`x - 1 = t => dx = dt`

Replacing the variable, yields:

`int (dx)/(x - 1) = int (dt)/t`

`int (dt)/t = ln|t| + c`

Replacing back `x - 1` for `t` , yields:

`int (dx)/(x - 1) = ln|x - 1| + c`

Hence, evaluating the indefinite integral, under the given conditions, yields `int (dx)/(x - 1) = ln|x - 1| + c.`

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