Integral of \( \displaystyle \frac{\left(2 x - 1\right)^{2} + 1}{\left(2 x - 2\right) \left(2 x - 1\right)} \)
Problem 4.259 · hard
Find \( \displaystyle \int \frac{\left(2 x - 1\right)^{2} + 1}{\left(2 x - 2\right) \left(2 x - 1\right)} \, dx \). (Omit the constant of integration.)
- \[ \int \frac{\left(2 x - 1\right)^{2} + 1}{\left(2 x - 2\right) \left(2 x - 1\right)}\, dx \]integral algebraStart with the integral of the given function. Factor out 2 from the first term in the denominator.✓ Proved
- \[ = \int \left(\frac{2 x - 1}{2 x - 2} + \frac{1}{\left(2 x - 2\right) \left(2 x - 1\right)}\right)\, dx \]linearity simplify algebraSplit the integrand into two parts. Cancel the common factor (2*x - 1) in the first term. Rewrite the numerator to facilitate division.✓ Proved
- \[ = \int \left(1 + \frac{1}{2 x - 2} + \frac{1}{\left(2 x - 2\right) \left(2 x - 1\right)}\right)\, dx \]long-divisionSimplify the first term using division.✓ Proved
- \[ = \int 1\, dx + \int \frac{1}{\left(2 x - 2\right) \left(2 x - 1\right)}\, dx + \int \frac{1}{2 x - 2}\, dx \]linearitySplit the integral into three parts.✓ Proved
- \[ = x + \int \frac{1}{\left(2 x - 2\right) \left(2 x - 1\right)}\, dx + \int \frac{1}{2 x - 2}\, dx \]antiderivativeIntegrate the constant term.✓ Proved
- \[ = x + \frac{\ln{\left(x - 1 \right)}}{2} + \int \frac{1}{\left(2 x - 2\right) \left(2 x - 1\right)}\, dx \]antiderivativeIntegrate the second term.✓ Proved
- \[ = x + \frac{\ln{\left(x - 1 \right)}}{2} + \int \left(- \frac{1}{2 x - 1} + \frac{1}{2 x - 2}\right)\, dx \]rewriteUse partial fraction decomposition on the remaining integrand.✓ Proved
- \[ = x + \frac{\ln{\left(x - 1 \right)}}{2} + \int \frac{1}{2 x - 2}\, dx - \int \frac{1}{2 x - 1}\, dx \]linearitySplit the integral of the partial fraction sum.≈ Checked numerically
- \[ = x + \ln{\left(x - 1 \right)} - \int \frac{1}{2 x - 1}\, dx \]antiderivative simplifyIntegrate the third term. Combine the log terms.✓ Proved
- \[ = x + \ln{\left(x - 1 \right)} - \frac{\ln{\left(2 x - 1 \right)}}{2} \]antiderivativeIntegrate the final term using substitution rule for 1/(ax+b).✓ Proved
Answer \( x + \ln{\left(x - 1 \right)} - \frac{\ln{\left(x - \frac{1}{2} \right)}}{2} + C \)
Lines: 13 proved, 2 checked numerically. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in 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 2*x - 1 = 0 undefined where 2*x - 2 = 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 2*x - 1 = 0 undefined where 2*x - 2 = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 2*x - 1 = 0 undefined where 2*x - 2 = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 2*x - 1 = 0 undefined where 2*x - 2 = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 2*x - 1 = 0 undefined where 2*x - 2 = 0 |
| 7 | ✓ Proved | sympy 1.14.0 | lines differ by the constant -log(2)/2 undefined where 2*x - 2 = 0 undefined where 2*x - 1 = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 2*x - 2 = 0 undefined where 2*x - 1 = 0 |
| 9 | ✓ Proved | sympy 1.14.0 | lines differ by the constant log(2)/2 undefined where 2*x - 2 = 0 undefined where 2*x - 1 = 0 log is undefined for non-positive arguments |
| 10 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 undefined where 2*x - 2 = 0 |
| 11 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left -log(x - 1/2)/2 + log(2*x - 1)/2 - log(2)/2; numeric agreement only, at 24 of 24 sampled points log is undefined for non-positive arguments undefined where 2*x - 2 = 0 undefined where 2*x - 1 = 0 |
| 12 | ✓ Proved | sympy 1.14.0 | lines differ by the constant log(2)/2 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 undefined where 2*x - 2 = 0 |
| 13 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| 14 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 log is undefined for non-positive arguments undefined where 2*x - 1 = 0 |
| answer | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: final line against the stated answer: simplify left log(x - 1/2)/2 - log(2*x - 1)/2; numeric agreement only, at 24 of 24 sampled points 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 (style) — Step 10 applies a partial‑fractions decomposition but labels it "rewrite", which is not an allowed rule name. The correct label should be "partial-fractions".qwen3.6:27b-mlx: fail (style) — Step 10 is labeled 'rewrite' but performs partial fraction decomposition, which should be labeled 'partial-fractions'. Step 14 is labeled 'antiderivative' but the note describes a substitution; while 'antiderivative' is acceptable for the result, the label 'substitution' would be more precise given the note, or the note should be removed to match the label. However, the primary defect is Step 10 mislabeling the specific algebraic technique.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (style) 2026-10-04 — Step 10 is labeled 'rewrite' but performs partial fraction decomposition, which should be labeled 'partial-fractions'. Step 14 is labeled 'antiderivative' but the note describes a substitution; while 'antiderivative' is acceptable for the result, the label 'substitution' would be more precise given the note, or the note should be removed to match the label. However, the primary defect is Step 10 mislabeling the specific algebraic technique.gpt-oss:20b: fail (style) 2026-10-04 — Step 10 applies a partial‑fractions decomposition but labels it "rewrite", which is not an allowed rule name. The correct label should be "partial-fractions".qwen3.6:27b-mlx: fail (style) 2026-10-04 — Step 10 applies partial fraction decomposition but is labeled 'rewrite'; the vocabulary includes 'partial-fractions' which is the correct label for this operation. Additionally, Step 14 is labeled 'antiderivative' but the note mentions 'substitution rule'; while 'antiderivative' is acceptable for the result, the note is slightly misleading about the mechanism, though the label itself is valid. The primary defect is the mislabeling of Step 10.gpt-oss:20b: fail (error) 2026-10-04 — Step 10 applies partial‑fraction decomposition incorrectly: the correct decomposition is 1/(2*(x-1)*(2*x-1)) = 1/2·1/(x-1) – 1/(2*x-1). The solution omits the 1/2 factor on the first term, so the subsequent integrals are based on a wrong integrand.
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 2026-10-04 with SymPy 1.14.0.