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# What is the integration of log ((1- x) / x) with limit 0 to 1?

**Solution:**

To find the value of this we will use \(\int_0^{2a}\)f (x) dx = \(\int_0^{2a}\) f (x - 2a) dx.

⇒ Let I =\(\int_0^{1}\) ( log ((1 - x) / x)dx ----- (1)

Using \(\int_0^{2a}\)f (x) dx = \(\int_0^{2a}\) f (x - 2a) dx let us evaluate \(\int_0^1f(x - 1) dx\)

When f(x) = log ((1 - x) / x), then f(x-1) = log ((1 - (x-1)) / (x-1))

= log(-x)/(x-1)= log (x/(1-x))

⇒ I =\(\int_0^{1}\)log (x / (1 - x)) ----- (2)

On adding both the equations, we get

⇒ I + I = ^{ }\(\int_0^{1}\)(log ( (1 - x) / x ) dx + \(\int_0^{1}\)(log (x / (1 - x)) dx

⇒ 2I = \(\int_0^{1}\) (log ((1 - x) / x × x / (1 - x)) dx

⇒ 2I = \(\int_0^{1}\) log 1

⇒ 0

Therefore, I = 0

Thus, the value of \(\int_0^{1}\) ( log ((1 - x) / x) dx is 0.

## What is the integration of log ((1- x) / x) with limit 0 to 1?

**Summary:**

The value of \(\int_0^1\) log ((1 - x) / x) dx is 0.

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