∫Calc Practice

Substitution with a given u

Problem 4.489 · hard

Evaluate \( \displaystyle \int x^{2} \left(x^{3} + 2\right)^{5}\, dx \) using the substitution \( \displaystyle u = x^{3} + 2 \).
  1. \[ \frac{d}{d x} \left(x^{3} + 2\right) = 3 x^{2} \]
    So du = 3x² dx.✓ Proved
  2. In terms of u the integral becomes ∫ u**5/3 du.
    Reviewed
  3. \[ \frac{d}{d u} \frac{u^{6}}{18} = \frac{u^{5}}{3} \]
    An antiderivative in u is u**6/18; differentiating confirms it.✓ Proved
  4. \[ \frac{\left(x^{3} + 2\right)^{6}}{18} \]
    Replace u by x**3 + 2.✓ Proved
  5. \[ \frac{d}{d x} \frac{\left(x^{3} + 2\right)^{6}}{18} = x^{2} \left(x^{3} + 2\right)^{5} \]
    Differentiating the answer returns the integrand.✓ Proved
Answer \( \frac{\left(x^{3} + 2\right)^{6}}{18} + C \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.

The full receipt
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxa sentence; read, not computed
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0a difference quotient of the answer matches the integrand at x = 0.4, 0.9, 1.3

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass — The solution correctly applies the substitution method, handles the constant factor from the differential, and verifies the result.
Every verdict on record (4)
  • gpt-oss:20b: pass 2026-10-06
  • qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly applies the substitution method, handles the constant factor from the differential, and verifies the result.
  • gpt-oss:20b: pass 2026-10-06
  • qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly applies the substitution method, handles the constant factor from the differential, and verifies the result.

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 generator:structured/indefinite_substitution, checked 2026-10-06 with SymPy 1.14.0.