Integral of \( \displaystyle \frac{e^{x}}{e^{x} + 1} \)
Problem 4.728 · medium
Find \( \displaystyle \int \frac{e^{x}}{e^{x} + 1} \, dx \). (Omit the constant of integration.)
- \[ \int \frac{e^{x}}{e^{x} + 1}\, dx \]integral rewriteStart with the integral of the given function. Rewrite the fraction using a negative exponent.✓ Proved
- \[ = \int \frac{1}{1 + e^{- x}}\, dx \]substitutionUse the substitution u = exp(x) + 1, noting that du = exp(x) dx, which is equivalent to rewriting the integrand as 1/(1 + exp(-x)).✓ Proved
- \[ = \ln{\left(e^{x} + 1 \right)} \]antiderivativeIntegrate the simplified expression.✓ Proved
Answer \( \ln{\left(e^{x} + 1 \right)} + C \)
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 | simplify(a - b) reduced to 0 undefined where exp(x) + 1 = 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where exp(x) + 1 = 0 undefined where 1 + exp(-x) = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 1 + exp(-x) = 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: fail (error) — Step 3 applies two rules at once: it rewrites the integrand as 1/(1+exp(-x)) and claims this comes from the substitution u=exp(x)+1, which is incorrect. The correct substitution would give Integral(1/(u), du)=log(u), but the integrand was not properly transformed. Thus the solution contains an algebraic mistake.qwen3.6:27b-mlx: fail (error) — Step 3 claims to use substitution but produces an algebraically incorrect integrand (exp(x)/(exp(x)+1) is not equal to 1/(1+exp(-x))). Furthermore, the step applies multiple operations (algebraic manipulation and variable substitution) simultaneously, violating the one-rule-per-step constraint.
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-09 — Step 3 applies two rules at once: it rewrites the integrand as 1/(1+exp(-x)) and claims this comes from the substitution u=exp(x)+1, which is incorrect. The correct substitution would give Integral(1/(u), du)=log(u), but the integrand was not properly transformed. Thus the solution contains an algebraic mistake.qwen3.6:27b-mlx: fail (error) 2026-10-09 — Step 3 claims to use substitution but produces an algebraically incorrect integrand (exp(x)/(exp(x)+1) is not equal to 1/(1+exp(-x))). Furthermore, the step applies multiple operations (algebraic manipulation and variable substitution) simultaneously, violating the one-rule-per-step constraint.gpt-oss:20b: fail (misleading) 2026-10-09 — Step 3 incorrectly labels the transformation as a substitution. The integrand is simply rewritten algebraically as 1/(1+exp(-x)); no substitution is performed, so the label is misleading.qwen3.6:27b-mlx: fail (error) 2026-10-09 — Step 3 claims to use substitution but the resulting integrand 1/(1 + exp(-x)) is not algebraically equal to the previous step's integrand exp(x)/(exp(x) + 1). The substitution logic described in the note (u = exp(x) + 1) does not lead to the expression shown in step 3, and the algebraic transformation is invalid.
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.