Substitution with a given u
Problem 4.617 · hard
Evaluate \( \displaystyle \int \frac{6 e^{3 x}}{e^{3 x} + 3}\, dx \) using the substitution \( \displaystyle u = e^{3 x} + 3 \).
- \[ \frac{d}{d x} \left(e^{3 x} + 3\right) = 3 e^{3 x} \]So du = 3e^(3x) dx.✓ Proved
- In terms of u the integral becomes ∫ 2/u du.Reviewed
- \[ \frac{d}{d u} 2 \ln{\left(u \right)} = \frac{2}{u} \]An antiderivative in u is 2*log(u); differentiating confirms it.✓ Proved
- \[ 2 \ln{\left(e^{3 x} + 3 \right)} \]Replace u by exp(3*x) + 3.✓ Proved
- \[ \frac{d}{d x} 2 \ln{\left(e^{3 x} + 3 \right)} = \frac{6 e^{3 x}}{e^{3 x} + 3} \]Differentiating the answer returns the integrand.✓ Proved
Answer \( 2 \ln{\left(e^{3 x} + 3 \right)} + C \)
✓ Nihil obstat Lines: 4 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ 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 | a difference quotient of the answer matches the integrand at x = 0.4, 0.9, 1.3 |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies the substitution method, handles the constant factor adjustment accurately, and verifies the result. No errors or misleading statements found.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly applies the substitution method, handles the constant factor adjustment accurately, and verifies the result. No errors or misleading statements found.gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly applies the substitution method, handles the constant factor appropriately, and verifies the result by differentiation.
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/indefinite_substitution, checked 2026-10-07 with SymPy 1.14.0.