`int_0^1log(1/x-1)dx` is equal to

Explanation

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

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

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

or,

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

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


`=>2I=int_0^1[log(x(1-x))/((1-x)x)]dx`
...