The value of $\int_0^1 [\log x - \log(1-x)] dx$ is

Given: $\int_0^1 [\log x - \log(1-x)] \, dx$

The integral can be split into two parts:

$\int_0^1 [\log x - \log(1-x)] \, dx = \int_0^1 \log x \, dx - \int_0^1 \log(1-x) \, dx$

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Consider the second integral: $\int_0^1 \log(1-x) \, dx$

Let $u = 1-x$

Then $du = -dx$

Change of limits:

• When $x = 0$, $u = 1$
• When $x = 1$, $u = 0$

Substituting:

$\int_0^1 \log(1-x) \, dx = \int_1^0 \log u \cdot (-du)$

$= -\int_1^0 \log u \, du$

$= \int_0^1 \log u \, du$

$= \int_0^1 \log x \, dx$

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Substituting back into the original expression:

$\int_0^1 [\log x - \log(1-x)] \, dx = \int_0^1 \log x \, dx - \int_0^1 \log(1-x) \, dx$

$= \int_0^1 \log x \, dx - \int_0^1 \log x \, dx$

$= 0$

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Therefore, $\int_0^1 [\log x - \log(1-x)] \, dx = 0$