Integral of \( \displaystyle \frac{\ln{\left(x + 1 \right)}}{x + 1} \)
Problem 4.1401 · medium Video
Find \( \displaystyle \int \frac{\ln{\left(x + 1 \right)}}{x + 1} \, dx \). (Omit the constant of integration.)
- \[ \int \frac{\ln{\left(x + 1 \right)}}{x + 1}\, dx \]integral rewriteStart with the given integral. Rewrite the fraction using a negative exponent.✓ Proved
- \[ = \frac{\ln{\left(x + 1 \right)}^{2}}{2} \]antiderivativeApply the power rule for integration via substitution.✓ Proved
Answer \( \frac{\ln{\left(x + 1 \right)}^{2}}{2} + C \)
Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | line 1 is the problem as stated |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 1 = 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where x + 1 = 0 |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| answer, a second way | ✓ Proved | sympy 1.14.0 | SymPy differentiated the stated antiderivative back to the integrand |
Reviewers
gpt-oss:20b: fail (error) — Step 3 applies both a substitution (u=log(x+1)) and an antiderivative (∫u du = u²/2) in a single line, violating the rule that each step must change only one thing.qwen3.6:27b-mlx: pass — The solution correctly identifies the integral and applies substitution implicitly within the antiderivative step. The label 'antiderivative' is appropriate for recognizing the form u du -> u^2/2, and the intermediate rewrite step is valid.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-31 — The solution correctly identifies the integral and applies substitution implicitly within the antiderivative step. The label 'antiderivative' is appropriate for recognizing the form u du -> u^2/2, and the intermediate rewrite step is valid.gpt-oss:20b: fail (error) 2026-10-31 — Step 3 applies both a substitution (u=log(x+1)) and an antiderivative (∫u du = u²/2) in a single line, violating the rule that each step must change only one thing.qwen3.6:27b-mlx: pass 2026-10-30 — The solution correctly identifies the integral and applies substitution implicitly within the antiderivative step. The label 'antiderivative' is appropriate for recognizing the form u du where u = log(x+1).gpt-oss:20b: pass 2026-10-30qwen3.6:27b-mlx: pass 2026-10-30 — The solution correctly identifies the integral and applies substitution implicitly within the antiderivative step. The label 'antiderivative' is appropriate for recognizing the form u du where u = log(x+1).gpt-oss:20b: pass 2026-10-30
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
gemma4:26b, checked 2026-10-31 with SymPy 1.14.0.