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Home›Calculus 2›Partial fractions›Problem 4.795

Integral of \( \displaystyle \frac{\left(2 x + 1\right)^{2} + 1}{2 x \left(2 x + 1\right)} \)

Problem 4.795 · hard

Find \( \displaystyle \int \frac{\left(2 x + 1\right)^{2} + 1}{2 x \left(2 x + 1\right)} \, dx \). (Omit the constant of integration.)
  1. \[ \int \left(\frac{2 x + 1}{2 x} + \frac{1}{2 x \left(2 x + 1\right)}\right)\, dx \]
    algebra simplifySplit the fraction using the numerator's terms. Cancel common terms in the first part of the integrand.✓ Proved
  2. \[ = \int \left(1 + \frac{1}{2 x} + \frac{1}{2 x \left(2 x + 1\right)}\right)\, dx \]
    algebraDivide the numerator by the denominator in the first term.✓ Proved
  3. \[ = \int 1\, dx + \int \frac{1}{2 x}\, dx + \int \frac{1}{2 x \left(2 x + 1\right)}\, dx \]
    linearityDistribute the integral across the sum.✓ Proved
  4. \[ = \int 1\, dx + \int \frac{1}{2 x}\, dx + \int \left(- \frac{1}{2 x + 1} + \frac{1}{2 x}\right)\, dx \]
    rewriteUse partial fraction decomposition on the third term.✓ Proved
  5. \[ = \int 1\, dx + 2 \int \frac{1}{2 x}\, dx - \int \frac{1}{2 x + 1}\, dx \]
    linearity simplifySplit the integral of the decomposed term. Combine the identical integral terms.≈ Checked numerically
  6. \[ = x + \ln{\left(2 x \right)} - \frac{\ln{\left(2 x + 1 \right)}}{2} \]
    antiderivative simplifyIntegrate each term separately. Simplify the coefficients.✓ Proved
  7. \[ = x + \ln{\left(2 x \right)} - \ln{\left(\sqrt{2 x + 1} \right)} \]
    rewriteRewrite the division by 2 as a square root inside the logarithm.✓ Proved
  8. \[ = x + \ln{\left(\frac{2 x}{\sqrt{2 x + 1}} \right)} \]
    simplifyUse logarithmic properties to combine the terms into a single logarithm.≈ Checked numerically
Answer \( x + \ln{\left(x \right)} - \frac{\ln{\left(x + \frac{1}{2} \right)}}{2} + C \)

Lines: 9 proved, 3 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
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0line 1 is the problem as stated
2✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
undefined where x = 0
undefined where 2*x + 1 = 0
3✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
undefined where x = 0
undefined where 2*x + 1 = 0
4✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
undefined where x = 0
undefined where 2*x + 1 = 0
5✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
undefined where x = 0
undefined where 2*x + 1 = 0
6≈ Checked numericallysympy 1.14.0sympy 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 x = 0
undefined where 2*x + 1 = 0
7✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
undefined where x = 0
undefined where 2*x + 1 = 0
8✓ Provedsympy 1.14.0lines differ by the constant -log(2)
undefined where x = 0
undefined where 2*x + 1 = 0
log is undefined for non-positive arguments
9✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
10✓ Provedsympy 1.14.0simplify(a - b) reduced to 0
log is undefined for non-positive arguments
11≈ Checked numericallysympy 1.14.0sympy 1.14.0: simplify left log(x) - log(x/sqrt(2*x + 1)) - log(2*x + 1)/2; numeric agreement only, at 24 of 24 sampled points
log is undefined for non-positive arguments
undefined where 2*x + 1 = 0
answer≈ Checked numericallysympy 1.14.0sympy 1.14.0: final line against the stated answer: simplify left -log(x) + log(x/sqrt(2*x + 1)) + log(x + 1/2)/2 + log(2); numeric agreement only, at 24 of 24 sampled points
log is undefined for non-positive arguments
answer, a second way✓ Provedsympy 1.14.0SymPy differentiated the stated antiderivative back to the integrand

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (misleading) — Step 10 incorrectly rewrites log(2*x + 1)/2 as log(sqrt(2*x + 1)), which is only valid when 2*x + 1 > 0. The correct identity is log(|2*x + 1|)/2 = log(sqrt(|2*x + 1|)). This omission of absolute values makes the solution invalid for domains where 2*x + 1 < 0.
Every verdict on record (4)
  • gpt-oss:20b: pass 2026-10-10
  • qwen3.6:27b-mlx: fail (misleading) 2026-10-10 — Step 10 incorrectly rewrites log(2*x + 1)/2 as log(sqrt(2*x + 1)), which is only valid when 2*x + 1 > 0. The correct identity is log(|2*x + 1|)/2 = log(sqrt(|2*x + 1|)). This omission of absolute values makes the solution invalid for domains where 2*x + 1 < 0.
  • gpt-oss:20b: pass 2026-10-10
  • qwen3.6:27b-mlx: fail (style) 2026-10-10 — [domain objection, downgraded to style] Step 10 is mathematically incorrect: log(2*x + 1)/2 is not equal to log(sqrt(2*x + 1)) without absolute values, and the transformation is unnecessary and confusing. Step 11 combines logs incorrectly by ignoring the domain restrictions and absolute values required for log(a) - log(b) = log(a/b). The final answer form is also inconsistent with the stated answer provided in the task description.

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 2026-10-10 with SymPy 1.14.0.