Evaluate the following integral:`int_0^1log(1/x-1)dx`

Explanation

Given:`int_0^1log(1/x-1)dx`

`I=int_0^1log(1/x-1)dx`

`I=int_0^1log((1-x)/x)dx....(i)`

`I=int_0^1log((1-(1-x))/(1-x))dx`

`I=int_0^1log(x/(1-x))dx.....(ii)`

As `int_0^a f(x)dx=int_0^a f(a-x)dx`

on adding (i) and (ii) we get

`2I=int_0^1 log ((1-x)/x)dx+int_0^1 log(x/(1-x))dx`

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