Integral of \( \displaystyle \frac{e^{3 x}}{e^{3 x} + 1} \)
Problem 4.880 · medium
Find \( \displaystyle \int \frac{e^{3 x}}{e^{3 x} + 1} \, dx \). (Omit the constant of integration.)
- \[ \int \frac{e^{3 x}}{e^{3 x} + 1}\, dx \]integral algebraStart with the integral of the given function. Rewrite the integrand by multiplying and dividing by 3.✓ Proved
- \[ = \frac{\ln{\left(e^{3 x} + 1 \right)}}{3} \]antiderivativeThe integral follows the form u/(du/dx) where u = exp(3*x) + 1.✓ Proved
Answer \( \frac{\ln{\left(e^{3 x} + 1 \right)}}{3} + C \)
✓ Nihil obstat 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. Reviewers found nothing wrong with the explanation.
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 undefined where exp(3*x) + 1 = 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where exp(3*x) + 1 = 0 log is undefined for non-positive arguments |
| 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: passqwen3.6:27b-mlx: pass
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-11gpt-oss:20b: pass 2026-10-11qwen3.6:27b-mlx: pass 2026-10-11gpt-oss:20b: pass 2026-10-11
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 2026-10-11 with SymPy 1.14.0.