Definite integrals by substitution
Problem 4.604 · medium
Evaluate \( \displaystyle \int_{0}^{1} \frac{5 e^{x}}{\left(e^{x} + 1\right)^{2}}\, dx \).
- Let u = exp(x) + 1; then du = exp(x) dx, which is in the integrand up to a constant.Reviewed
- \[ \left. e^{x} + 1 \right|_{\substack{ x=0 }} = 2 \]The lower limit in u.✓ Proved
- \[ \left. e^{x} + 1 \right|_{\substack{ x=1 }} = 1 + e \]The upper limit in u.✓ Proved
- \[ \int\limits_{2}^{1 + e} \frac{5}{u^{2}}\, du = \frac{-5 + 5 e}{2 + 2 e} \]Integrate in u between the new limits; no back-substitution needed.✓ Proved
Answer \( \frac{5 \left(-1 + e\right)}{2 \left(1 + e\right)} \)
✓ Nihil obstat Lines: 3 proved, 1 reviewed. 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | numerical quadrature of the original integral |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly applies u-substitution, accurately transforms the limits of integration, and computes the definite integral correctly.
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
generator:structured/definite_substitution, checked 2026-10-07 with SymPy 1.14.0.