Integral of \( \displaystyle \frac{\left(3 x + 1\right)^{2} + 1}{3 x \left(3 x + 1\right)} \)
Problem 4.792 · hard
Find \( \displaystyle \int \frac{\left(3 x + 1\right)^{2} + 1}{3 x \left(3 x + 1\right)} \, dx \). (Omit the constant of integration.)
- \[ \int \frac{\left(3 x + 1\right)^{2} + 1}{3 x \left(3 x + 1\right)}\, dx \]integralStart with the integral of the given function.✓ Proved
- \[ = \int \left(\frac{3 x + 1}{3 x} + \frac{1}{3 x \left(3 x + 1\right)}\right)\, dx \]linearity simplifySplit the integrand into two parts. Simplify the first term by canceling (3*x + 1).✓ Proved
- \[ = \int \left(1 + \frac{1}{3 x} + \frac{1}{3 x \left(3 x + 1\right)}\right)\, dx \]simplifySplit the first term into two parts.✓ Proved
- \[ = \int \left(1 - \frac{1}{3 x + 1} + \frac{2}{3 x}\right)\, dx \]partial-fractions simplifyApply partial fraction decomposition to the second term. Combine the like terms.✓ Proved
- \[ = x + \frac{2 \ln{\left(x \right)}}{3} - \frac{\ln{\left(3 x + 1 \right)}}{3} \]antiderivative simplifyIntegrate each term separately. Final simplified form of the antiderivative.≈ Checked numerically
Answer \( x + \frac{2 \ln{\left(x \right)}}{3} - \frac{\ln{\left(x + \frac{1}{3} \right)}}{3} + C \)
✓ Nihil obstat Lines: 7 proved, 2 checked numerically. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
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 x = 0 undefined where 3*x + 1 = 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where x = 0 undefined where 3*x + 1 = 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where x = 0 undefined where 3*x + 1 = 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where x = 0 undefined where 3*x + 1 = 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 0 undefined where x = 0 undefined where 3*x + 1 = 0 |
| 7 | ≈ Checked numerically | sympy 1.14.0 | sympy 1.14.0: simplify left -log(x + 1/3)/3 + log(3*x + 1)/3; numeric agreement only, at 24 of 24 sampled points undefined where x = 0 undefined where 3*x + 1 = 0 log is undefined for non-positive arguments |
| 8 | ✓ Proved | sympy 1.14.0 | simplify(a - b) reduced to 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 + 1/3)/3 - log(3*x + 1)/3; 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: pass — The solution correctly applies linearity, simplification, and partial fraction decomposition. The final antiderivative is algebraically equivalent to the stated answer, as log(3x+1) and log(x+1/3) differ only by a constant.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly applies linearity, simplification, and partial fraction decomposition. The final antiderivative is algebraically equivalent to the stated answer, as log(3x+1) and log(x+1/3) differ only by a constant.qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly applies linearity, simplification, and partial fraction decomposition in distinct steps. The final antiderivative is algebraically equivalent to the stated answer, as log(3x+1) differs from log(x+1/3) only by a constant (log 3), which is absorbed into the constant of integration.gpt-oss:20b: pass 2026-10-10
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.