Question

`int_0^1 log(1/x - 1) "dx"` = ______.

Options

• 0

• 1

• `1/2`

• `1/3`

Answer 1

`int_0^1 log(1/x - 1) "dx"` = 0.

Explanation:

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

∴ I = `int_0^1 log ((1 - x)/x)`dx

`= int_0^1 [(1 - (1 - x))/(1 - x)] "dx"   ...[because int_0^"a" "f"(x) "dx" = int_0^"a" "f"("a - x")"dx"]`

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

Adding (i) and (ii), we get

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

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

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

`= int_0^1 log 1 "dx"`

∴ 2I = `int_0^1` 0 dx 

∴ I = 0