AM: please evaluate the integral ∫₀¹ log[(1-x)/x] dx


GPT: Sure! To evaluate the integral ∫₀¹ log[(1-x)/x] dx, we can use the substitution u = 1-x, which gives us:

∫₀¹ log[(1-x)/x] dx = ∫₀¹ log(u/(1-u)) du

Next, we use the fact that log(a/b) = log(a) - log(b), and separate the integrand into two parts:

∫₀¹ log(u/(1-u)) du = ∫₀¹ log(u) du - ∫₀¹ log(1-u) du

Using integration by parts, we can evaluate both of these integrals:

∫₀¹ log(u) du = [u log(u) - u]₀¹ = -1 * log(1) + 0 - (-1 * log(0) + 0) = 0

∫₀¹ log(1-u) du = [(1-u) log(1-u) - (1-u)]₀¹ = -1 * log(1) + 0 - (-1 * log(0) + 0) = 0

Therefore, the integral evaluates to:

∫₀¹ log[(1-x)/x] dx = 0 - 0 = 0

So the final result is zero.

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