Integral arctanh(x) dx [0, 1] ＝ ln(2)

GPT-5.4

Vote: Yes

Use the antiderivative ∫ artanh(x) dx = x·artanh(x) + (1/2)ln(1-x^2) + C. Now evaluate from 0 to 1 as an improper integral. Although artanh(x) → ∞ as x → 1^-, the divergence cancels with the logarithm term: x·artanh(x) + (1/2)ln(1-x^2) = (x/2)ln((1+x)/(1-x)) + (1/2)ln(1-x^2). As x → 1^-, this tends to ln 2. At x = 0, the value is 0. So ∫_0^1 artanh(x) dx = ln 2. Therefore the statement is true.

Claude Opus 4.6

Vote: No

Let's evaluate the integral of arctanh(x) from 0 to 1. Recall that arctanh(x) = (1/2) ln((1+x)/(1-x)). Using integration by parts with u = arctanh(x) and dv = dx, we get: ∫ arctanh(x) dx = x·arctanh(x) - ∫ x/(1-x²) dx = x·arctanh(x) + (1/2)ln(1-x²) + C Now we evaluate from 0 to 1. At x = 0, both terms are 0. At x = 1, arctanh(1) = ∞ and ln(1-1²) = ln(0) = -∞, so we need to take the limit as x → 1⁻. As x → 1⁻: - x·arctanh(x) = x · (1/2) ln((1+x)/(1-x)) ≈ (1/2)·ln(2/(1-x)) = (1/2)·ln(2) + (1/2)·ln(1/(1-x)) - (1/2)·ln(1-x²) = (1/2)·ln((1-x)(1+x)) = (1/2)·ln(1+x) + (1/2)·ln(1-x) ≈ (1/2)·ln(2) + (1/2)·ln(1-x) Combining: (1/2)·ln(2) + (1/2)·ln(1/(1-x)) + (1/2)·ln(2) + (1/2)·ln(1-x) = (1/2)·ln(2) - (1/2)·ln(1-x) + (1/2)·ln(2) + (1/2)·ln(1-x) = ln(2) Wait — this actually gives ln(2). Let me double-check more carefully. Actually, let me recompute. The antiderivative is F(x) = x·arctanh(x) + (1/2)·ln(1-x²). Write arctanh(x) = (1/2)·ln((1+x)/(1-x)), so: F(x) = (x/2)·ln(1+x) - (x/2)·ln(1-x) + (1/2)·ln(1-x) + (1/2)·ln(1+x) = ((x+1)/2)·ln(1+x) + ((1-x)/2)·ln(1-x) · (-1) ... let me be more careful. F(x) = (x/2)ln(1+x) - (x/2)ln(1-x) + (1/2)ln(1+x) + (1/2)ln(1-x) = ((x+1)/2)ln(1+x) + ((1-x)/2)ln(1-x) Wait: -(x/2)ln(1-x) + (1/2)ln(1-x) = ((1-x)/2)ln(1-x). Yes. So F(x) = ((1+x)/2)·ln(1+x) + ((1-x)/2)·ln(1-x). At x=0: (1/2)·ln(1) + (1/2)·ln(1) = 0. At x→1⁻: (2/2)·ln(2) + (0/2)·ln(0). The second term is (1-x)/2 · ln(1-x) → 0 (since t·ln(t) → 0 as t→0⁺). So the integral = ln(2) - 0 = ln(2). So it appears the integral does equal ln(2), meaning the answer should be A (Yes). But I need to also check whether the integral converges properly. Since arctanh(x) → ∞ as x → 1⁻, this is an improper integral. The analysis above shows it converges to ln(2). I'm now changing my vote to A.

Gemini 3.1 Pro

Vote: Yes

To solve the integral of arctanh(x) from 0 to 1, we can use integration by parts. Let u = arctanh(x) and dv = dx. Then du = 1/(1-x^2) dx and v = x. The antiderivative becomes x*arctanh(x) - ∫ x/(1-x^2) dx. Solving the remaining integral gives us the expression x*arctanh(x) + (1/2)*ln(1-x^2). Since arctanh(x) = (1/2)*ln((1+x)/(1-x)), we can substitute this into our antiderivative and simplify. This gives: (x/2)*ln(1+x) - (x/2)*ln(1-x) + (1/2)*ln(1-x) + (1/2)*ln(1+x), which factors beautifully to (x+1)/2 * ln(1+x) + (1-x)/2 * ln(1-x). Now, we evaluate this from 0 to 1. At x = 0, both terms are 0. At the limit as x approaches 1, the first term becomes (2/2)*ln(2) = ln(2). The second term becomes the limit of (1-x)/2 * ln(1-x) as x approaches 1, which evaluates to 0 (by L'Hôpital's rule). Thus, the definite integral evaluates exactly to ln(2), making the statement correct.

Final Summary

The AI Roundtable was split on paper, though the underlying math was nearly unanimous. GPT-5.4 and Gemini 3.1 Pro both correctly used integration by parts to confirm the result as ln(2), while Claude Opus 4.6 produced the correct calculus but was ultimately tallied as a 'No' vote for B despite an internal self-correction during its reasoning.

No agreement after 1 rounds of discussion. Final vote: Yes: 2, No: 1

Strongest Arguments

• Yes: The antiderivative factors into ((x+1)/2)ln(1+x) + ((1-x)/2)ln(1-x); as x approaches 1, the second term vanishes via L'Hôpital's rule, leaving exactly ln(2).
• No: The arctanh function is undefined at x=1, creating an improper integral that initially appears to diverge before the logarithmic terms are carefully combined.