Integral of \( \displaystyle \frac{6 x + 2}{- 2 x + \left(2 x + 1\right)^{2} - 1} \)
Problem 4.872 · hard
Find \( \displaystyle \int \frac{6 x + 2}{- 2 x + \left(2 x + 1\right)^{2} - 1} \, dx \). (Omit the constant of integration.)
- \[ \int \frac{6 x + 2}{- 2 x + \left(2 x + 1\right)^{2} - 1}\, dx \]integralStart with the integral of the given function.✓ Proved
- \[ = \int \frac{6 x + 2}{4 x^{2} + 2 x}\, dx \]algebra simplifyExpand the squared term. Combine like terms in the denominator.✓ Proved
- \[ = \int \frac{6 x + 2}{2 x \left(2 x + 1\right)}\, dx \]algebraFactor the denominator.✓ Proved
- \[ = \int \frac{3 x + 1}{x \left(2 x + 1\right)}\, dx \]algebraCancel the common factor of 2.✓ Proved
- \[ = \int \left(\frac{1}{2 x + 1} + \frac{1}{x}\right)\, dx \]rewriteDecompose the integrand using partial fraction decomposition.✓ Proved
- \[ = \int \frac{1}{x}\, dx + \int \frac{1}{2 x + 1}\, dx \]linearitySplit the integral into two parts.≈ Checked numerically
- \[ = \ln{\left(x \right)} + 0.5 \ln{\left(2 x + 1 \right)} \]antiderivativeEvaluate the integrals.✓ Proved
Answer \( \ln{\left(x \right)} + \frac{\ln{\left(x + \frac{1}{2} \right)}}{2} + C \)
Lines: 7 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 + (2*x + 1)**2 - 1 = 0 undefined where 4*x**2 + 2*x = 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 4*x**2 + 2*x = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 4*x**2 + 2*x = 0 undefined where 2*x + 1 = 0 undefined where x = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 2*x + 1 = 0 undefined where x = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 2*x + 1 = 0 undefined where x = 0 |
| 7 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left log(x + 1/2)/2 - log(2*x + 1)/2; numeric agreement only, at 24 of 24 sampled points undefined where 2*x + 1 = 0 undefined where x = 0 |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where 2*x + 1 = 0 undefined where x = 0 log is undefined for non-positive arguments |
| answer | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: final line against the stated answer: simplify left -log(x + 1/2)/2 + 0.5*log(2*x + 1); 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: passqwen3.6:27b-mlx: fail (error) — Step 6 claims to use 'rewrite' for partial fraction decomposition, but the label 'partial-fractions' exists in the vocabulary and is the correct specific label. More critically, the decomposition in Step 6 is mathematically incorrect: (3x+1)/(x(2x+1)) does not equal 1/x + 1/(2x+1). The correct decomposition is 1/x + 1/(2x+1) is false; actually (3x+1)/(x(2x+1)) = A/x + B/(2x+1) => 3x+1 = A(2x+1) + Bx. Let x=0 => 1=A. Let x=-1/2 => -1/2 = -B/2 => B=1. So it is 1/x + 1/(2x+1). Wait, let me re-check. 1/x + 1/(2x+1) = (2x+1 + x)/(x(2x+1)) = (3x+1)/(x(2x+1)). The algebra is correct. However, the label 'rewrite' is used for partial fractions. The instructions say 'partial-fractions' is in the vocabulary. Using 'rewrite' instead of 'partial-fractions' is a labeling defect. Is it an error or style? The prompt says 'A label is correct if it names a rule the step legitimately applied.' Partial fractions is a specific algebraic manipulation. 'Rewrite' is vague. But usually, if a specific label exists, using a generic one might be considered a style issue or a defect depending on strictness. Let's look closer. Step 6 applies partial fractions. The label is 'rewrite'. The vocabulary includes 'partial-fractions'. Using 'rewrite' when 'partial-fractions' is available and more precise is likely a style defect or a minor error in labeling. However, look at Step 8. The result is log(x) + 0.5*log(2*x + 1). The stated answer is log(x) + log(x + 1/2)/2. These are equivalent up to a constant (log(2) is constant). The prompt says 'constant of integration is omitted'. So the answer form is acceptable. The main issue is the label in Step 6. Is 'rewrite' acceptable for partial fractions? The prompt says 'Each step names its rule from this fixed vocabulary... A label is correct if it names a rule the step legitimately applied.' Partial fraction decomposition is a specific algebraic technique. 'Rewrite' is very broad. If 'partial-fractions' is in the list, it should be used. Using 'rewrite' is imprecise. Is it a 'fail'? The prompt says 'Use fail only for a real defect.' A labeling imprecision is often 'style'. But wait, look at Step 4 and 5. Step 4 factors denominator. Step 5 cancels 2. Step 6 does PFD. The label 'rewrite' is used. In many contexts, 'rewrite' is a catch-all for algebraic manipulation that doesn't fit other specific labels, but 'partial-fractions' is explicitly listed. This suggests 'partial-fractions' is the intended label for this operation. Using 'rewrite' is a defect in labeling. Is it 'error', 'misleading', or 'style'? It's not mathematically wrong. It's not misleading about the math. It's a style/labeling issue. However, often in these adversarial reviews, using the wrong label from a fixed vocabulary is considered a defect. Let's check if there are other errors. Step 2: Expand (2x+1)^2 = 4x^2+4x+1. Correct. Step 3: 4x^2+4x+1-2x-1 = 4x^2+2x. Correct. Step 4: Factor 4x^2+2x = 2x(2x+1). Numerator 6x+2 = 2(3x+1). Correct. Step 5: Cancel 2. Correct. Step 6: PFD. Correct math. Label 'rewrite'. Step 7: Linearity. Correct. Step 8: Antiderivative. Correct. The only potential defect is the label in Step 6. If I must choose between pass and fail, and the label is technically from the vocabulary but imprecise, is it a fail? The prompt says 'A label is correct if it names a rule the step legitimately applied.' Did the step apply 'rewrite'? Yes, it rewrote the expression. Did it apply 'partial-fractions'? Yes. Since 'partial-fractions' is in the vocabulary, it is the *better* label. But is 'rewrite' *incorrect*? The prompt says 'no other' labels exist. It doesn't say 'use the most specific label'. It says 'names its rule'. If 'rewrite' is a valid rule name for algebraic rewriting, it might be acceptable. However, usually, if a specific label like 'partial-fractions' exists, using 'rewrite' is considered a failure to use the correct specific label. Let's look at the severity options. 'style' is for 'naming and granularity complaints'. This fits 'style'. But the verdict must be 'pass' or 'fail'. 'Use fail only for a real defect.' Is a suboptimal label a real defect? In strict grading, yes. In loose grading, no. Given the adversarial nature, I will flag it. But wait, is there a bigger error? Step 8 result: 0.5*log(2x+1). Stated answer: 0.5*log(x+0.5). 0.5*log(2x+1) = 0.5*log(2(x+0.5)) = 0.5*log(2) + 0.5*log(x+0.5). The constant 0.5*log(2) is omitted. This is allowed. So the math is fine. The label is the only issue. I will mark it as fail with severity style because the specific label 'partial-fractions' was available and appropriate.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-11 — Step 6 claims to use 'rewrite' for partial fraction decomposition, but the label 'partial-fractions' exists in the vocabulary and is the correct specific label. More critically, the decomposition in Step 6 is mathematically incorrect: (3x+1)/(x(2x+1)) does not equal 1/x + 1/(2x+1). The correct decomposition is 1/x + 1/(2x+1) is false; actually (3x+1)/(x(2x+1)) = A/x + B/(2x+1) => 3x+1 = A(2x+1) + Bx. Let x=0 => 1=A. Let x=-1/2 => -1/2 = -B/2 => B=1. So it is 1/x + 1/(2x+1). Wait, let me re-check. 1/x + 1/(2x+1) = (2x+1 + x)/(x(2x+1)) = (3x+1)/(x(2x+1)). The algebra is correct. However, the label 'rewrite' is used for partial fractions. The instructions say 'partial-fractions' is in the vocabulary. Using 'rewrite' instead of 'partial-fractions' is a labeling defect. Is it an error or style? The prompt says 'A label is correct if it names a rule the step legitimately applied.' Partial fractions is a specific algebraic manipulation. 'Rewrite' is vague. But usually, if a specific label exists, using a generic one might be considered a style issue or a defect depending on strictness. Let's look closer. Step 6 applies partial fractions. The label is 'rewrite'. The vocabulary includes 'partial-fractions'. Using 'rewrite' when 'partial-fractions' is available and more precise is likely a style defect or a minor error in labeling. However, look at Step 8. The result is log(x) + 0.5*log(2*x + 1). The stated answer is log(x) + log(x + 1/2)/2. These are equivalent up to a constant (log(2) is constant). The prompt says 'constant of integration is omitted'. So the answer form is acceptable. The main issue is the label in Step 6. Is 'rewrite' acceptable for partial fractions? The prompt says 'Each step names its rule from this fixed vocabulary... A label is correct if it names a rule the step legitimately applied.' Partial fraction decomposition is a specific algebraic technique. 'Rewrite' is very broad. If 'partial-fractions' is in the list, it should be used. Using 'rewrite' is imprecise. Is it a 'fail'? The prompt says 'Use fail only for a real defect.' A labeling imprecision is often 'style'. But wait, look at Step 4 and 5. Step 4 factors denominator. Step 5 cancels 2. Step 6 does PFD. The label 'rewrite' is used. In many contexts, 'rewrite' is a catch-all for algebraic manipulation that doesn't fit other specific labels, but 'partial-fractions' is explicitly listed. This suggests 'partial-fractions' is the intended label for this operation. Using 'rewrite' is a defect in labeling. Is it 'error', 'misleading', or 'style'? It's not mathematically wrong. It's not misleading about the math. It's a style/labeling issue. However, often in these adversarial reviews, using the wrong label from a fixed vocabulary is considered a defect. Let's check if there are other errors. Step 2: Expand (2x+1)^2 = 4x^2+4x+1. Correct. Step 3: 4x^2+4x+1-2x-1 = 4x^2+2x. Correct. Step 4: Factor 4x^2+2x = 2x(2x+1). Numerator 6x+2 = 2(3x+1). Correct. Step 5: Cancel 2. Correct. Step 6: PFD. Correct math. Label 'rewrite'. Step 7: Linearity. Correct. Step 8: Antiderivative. Correct. The only potential defect is the label in Step 6. If I must choose between pass and fail, and the label is technically from the vocabulary but imprecise, is it a fail? The prompt says 'A label is correct if it names a rule the step legitimately applied.' Did the step apply 'rewrite'? Yes, it rewrote the expression. Did it apply 'partial-fractions'? Yes. Since 'partial-fractions' is in the vocabulary, it is the *better* label. But is 'rewrite' *incorrect*? The prompt says 'no other' labels exist. It doesn't say 'use the most specific label'. It says 'names its rule'. If 'rewrite' is a valid rule name for algebraic rewriting, it might be acceptable. However, usually, if a specific label like 'partial-fractions' exists, using 'rewrite' is considered a failure to use the correct specific label. Let's look at the severity options. 'style' is for 'naming and granularity complaints'. This fits 'style'. But the verdict must be 'pass' or 'fail'. 'Use fail only for a real defect.' Is a suboptimal label a real defect? In strict grading, yes. In loose grading, no. Given the adversarial nature, I will flag it. But wait, is there a bigger error? Step 8 result: 0.5*log(2x+1). Stated answer: 0.5*log(x+0.5). 0.5*log(2x+1) = 0.5*log(2(x+0.5)) = 0.5*log(2) + 0.5*log(x+0.5). The constant 0.5*log(2) is omitted. This is allowed. So the math is fine. The label is the only issue. I will mark it as fail with severity style because the specific label 'partial-fractions' was available and appropriate.gpt-oss:20b: pass 2026-10-11qwen3.6:27b-mlx: pass 2026-10-11 — The solution correctly simplifies the integrand, applies partial fraction decomposition, and evaluates the resulting integrals. The final answer is algebraically equivalent to the stated answer.gpt-oss:20b: pass 2026-10-11
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 2026-10-11 with SymPy 1.14.0.