Question

Evaluate the following using properties of definite integral:

`int_0^1 log (1/x - 1)  "d"x`

Answer 1

Let I = `int_0^1 log (1/x - 1)  "d"x`

I = `int_0^1 log ((1 - x)/x)  "d"x`   ........(1)

Using the property

`int_0^"a" "f"(x)  "d"x = int_0^"a"  "f"("a" -x)  "d"x`

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

= `int_0^1 log((1 - (1 - x))/(1 - x))  "d"x`

= `int_0^1 log(x/(1 - x))  "d"x`

Adding (1) and (2)

I + I = `int_0^1 log((1 - x)/x)  "d"x + int_0^1 log (x/(1 - x ))  "d"x`

2I = `int_0^1 log [((1 - x))/x xx x/((1 - x))]  "d"x`

2I = `int_0^1 log(1)  "d"x` = 0 

∴ I = 0