Integral of \( \displaystyle \frac{e^{x + 1}}{e^{x + 1} + 1} \)
Problem 4.734 · medium
Find \( \displaystyle \int \frac{e^{x + 1}}{e^{x + 1} + 1} \, dx \). (Omit the constant of integration.)
- \[ \int \frac{e^{x + 1}}{e^{x + 1} + 1}\, dx \]integralStart with the integral of the given function.✓ Proved
- \[ = \int \frac{e e^{x}}{e e^{x} + 1}\, dx \]algebraUse the property exp(a+b) = exp(a)exp(b).✓ Proved
- \[ = \int \frac{e^{x}}{e^{x} + e^{-1}}\, dx \]algebraDivide numerator and denominator by exp(1).✓ Proved
- \[ = \ln{\left(e^{x} + e^{-1} \right)} \]substitution simplifyLet u = exp(x) + exp(-1), then du = exp(x)dx. The integral of the substitution terms cancels out.✓ Proved
Answer \( \ln{\left(e^{x + 1} + 1 \right)} + C \)
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. A reviewer's objection to the explanation stands; see the receipt.
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 | lines differ by the constant 1 undefined where exp(x + 1) + 1 = 0 undefined where E*exp(x) + 1 = 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where E*exp(x) + 1 = 0 undefined where exp(x) + exp(-1) = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where exp(x) + exp(-1) = 0 log is undefined for non-positive arguments |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments |
| answer | ✓ Proved | sympy 1.14.0 | final line against the stated answer: lines differ by the constant -1 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: inconclusive — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"Step 4 applies two rules at once: it performs the substitution \(u=e^x+e^{-1}\) **and** evaluates the resulting integral \(\int du/u\). The step is incorrqwen3.6:27b-mlx: fail (error) — Step 4 is mathematically nonsensical; it introduces two integral terms that cancel each other out while simultaneously asserting the result of the substitution, which is not a valid application of the substitution rule. The substitution rule should transform the integral into a new form (e.g., Integral(1/u, u)) rather than adding and subtracting the original integral.
Every verdict on record (4)
gpt-oss:20b: inconclusive 2026-10-09 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"Step 4 applies two rules at once: it performs the substitution \(u=e^x+e^{-1}\) **and** evaluates the resulting integral \(\int du/u\). The step is incorrqwen3.6:27b-mlx: fail (error) 2026-10-09 — Step 4 is mathematically nonsensical; it introduces two integral terms that cancel each other out while simultaneously asserting the result of the substitution, which is not a valid application of the substitution rule. The substitution rule should transform the integral into a new form (e.g., Integral(1/u, u)) rather than adding and subtracting the original integral.gpt-oss:20b: fail (error) 2026-10-09 — Step 4 applies two operations at once: it both performs a substitution and simultaneously rewrites the integral as a difference of two identical integrals plus a log term. This is not a single valid rule application and the resulting expression is incorrect. The correct antiderivative is log(exp(x+1)+1) without the extra cancellation step.qwen3.6:27b-mlx: fail (error) 2026-10-09 — Step 4 is logically incoherent; it introduces a subtraction of the same integral and adds the result simultaneously, which is not a valid application of the substitution rule. The substitution rule should transform the integral into a new form (e.g., Integral(1/u, u)), not create an identity like I - I + F.
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 2026-10-09 with SymPy 1.14.0.