Mastering Integration: Solving a Hybrid Substitution & Parts Problem (PDFjoiner 4-107) | 掌握积分：求解代换与分部混合题 (PDFjoiner 4-107)

📚 Mastering Integration: Solving a Hybrid Substitution & Parts Problem (PDFjoiner 4-107) | 掌握积分：求解代换与分部混合题 (PDFjoiner 4-107)

pot when a substitution and integration by parts must be used together. The exercise labelled 'PDFjoiner (4)-107' provides a perfect illustration of this blend. It asks us to find ∫ x³ e^x² dx – an integral that looks intimidating but unravels beautifully once you pick the right strategy. This article will guide you through the complete solution, explain the reasoning behind each choice, and strengthen your skills for mixed-method integration problems.

A-Level 阶段的积分往往不再只是单一套用公式。在 Edexcel 纯数考纲中，题目常要求你识别什么时候需要同时使用代换法和分部积分法。编号为 'PDFjoiner (4)-107' 的练习正是这种结合的最佳范例。它要求我们求 ∫ x³ e^x² dx —— 这个积分乍看令人畏惧，但选对策略后就能迎刃而解。本文将带你走完整个求解过程，解析每一步选择的理由，并强化你面对混合积分题时的解题能力。

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1. Setting the Scene: The PDFjoiner (4)-107 Integral | 背景设定：PDFjoiner (4)-107 积分题

The problem states: Find ∫ x³ e^x² dx. At first glance, you might try integration by parts directly, but the product of x³ and e^x² is not a simple power times exponential since the exponential's derivative yields an extra 2x factor. A pure substitution also struggles because of the cubic term. The solution lies in a strategic substitution that reduces the integral to a form where parts can be applied cleanly.

题目给出：计算 ∫ x³ e^x² dx。乍看之下，你也许想直接使用分部积分法，但 x³ 与 e^x² 的乘积不是简单的幂函数乘指数函数，因为指数部分的导数会多出一个 2x 因子。单纯使用代换法也因三次项而难以处理。解决办法是先作一个巧妙的代换，将积分化为能干净地套用分部积分法的形式。

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2. Recognising the Inner Function for Substitution | 识别代换的内部函数

Look at the exponent x². Its derivative is 2x, which almost appears in the integrand – uting this into the integral will cancel one power of x and leave a manageable expression.

观察指数部分 x²。它的导数是 2x，而我们的被积函数中有 x³，里面正好含有一个 x 因子。常用的技巧是设 u = x²。那么 du/dx = 2x，因此 dx = du/(2x)。将此代换代入积分后，会消去一个 x 的幂次，留下一个容易处理的表达式。

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3. Executing the Substitution | 实施代换

Let u = x². Then du = 2x dx, and we rewrite the integral as ∫ x³ e^x² dx = ∫ x² · e^x² · x dx. Since x² = u and x dx = du/2, the integral becomes ∫ u · e^u · (du/2) = (1/2) ∫ u e^u du. The apparently messy original now turns into a basic product: a linear term u multiplied by e^u, waiting for integration by parts.

令 u = x²。那么 du = 2x dx，我们将积分改写为 ∫ x³ e^x² dx = ∫ x² · e^x² · x dx。由于 x² = u 且 x dx = du/2，积分变为 ∫ u · e^u · (du/2) = (1/2) ∫ u e^u du。原本看似复杂的原积分，现在转化成了一个基本乘积：线性项 u 乘以 e^u，只需用分部积分法即可求解。

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4. Preparing for Integration by Parts | 为分部积分做准备

We now focus on I = ∫ u e^u du. Integration by parts follows the formula ∫ v dw = v w − ∫ w dv. A good choice here is to differentiate the polynomial u and integrate the exponential e^u, because differentiating u reduces its degree. So set v = u and dw = e^u du. Then dv = du and w = e^u.

我们现在专注于 I = ∫ u e^u du。分部积分公式为 ∫ v dw = v w − ∫ w dv。在这里，一个上佳的选择是对多项式 u 进行微分，而对指数 e^u 进行积分，因为对 u 微分能降低其次数。因此设 v = u，dw = e^u du。于是得到 dv = du，w = e^u。

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5. Applying the Parts Formula | 应用分部积分公式

Substituting into the parts formula: I = u · e^u − ∫ e^u du. The remaining integral is simply e^u. Thus I = u e^u − e^u + C₁, where C₁ is a constant of integration. Factorising gives I = e^u (u − 1) + C₁. This step is very standard and confirms why the 'differentiate polynomial, integrate exponential' priority works so cleanly.

代入分部积分公式：I = u · e^u − ∫ e^u du。剩下的积分就直接是 e^u。因此 I = u e^u − e^u + C₁，其中 C₁ 为积分常数。因式分解后得到 I = e^u (u − 1) + C₁。这一步非常标准，也印证了'先微分多项式、再积分指数'的优先顺序为何能处理得如此干净。

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6. Returning to the Original Variable | 回到原变量

Recall that our original integral J = ∫ x³ e^x² dx = (1/2) I. Substitute back u = x²: J = (1/2) [ e^x² (x² − 1) ] + C, where C = C₁/2. Writing it neatly: ∫ x³ e^x² dx = ½ e^x² (x² − 1) + C. This is the final answer. Always remember to replace u thoroughly and adjust the constant.

回顾一下，我们的原积分 J = ∫ x³ e^x² dx = (1/2) I。将 u = x² 代回：J = (1/2) [ e^x² (x² − 1) ] + C，其中 C = C₁/2。整洁地写出结果：∫ x³ e^x² dx = ½ e^x² (x² − 1) + C。这就是最终答案。务必记得完整地代回 u 并调整常数。

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7. Verification by Differentiation | 通过微分验证

A-Level examiners love candidates who check their work. Differentiate the result y = ½ e^x²(x² − 1) using the product rule. Let f = ½ e^x² and g = (x² − 1). Then f ' = ½ e^x² · 2x = x e^x², and g' = 2x. So dy/dx = f ' g + f g' = x e^x² (x² − 1) + ½ e^x² · 2x = x e^x²(x² − 1) + x e^x² = x e^x² [x² − 1 + 1] = x³ e^x², which matches the original integrand perfectly.

A-Level 考官非常欣赏那些会验算的考生。对结果 y = ½ e^x²(x² − 1) 用乘法法则求导。令 f = ½ e^x²，g = (x² − 1)。那么 f ' = ½ e^x² · 2x = x e^x²，g' = 2x。所以 dy/dx = f ' g + f g' = x e^x² (x² − 1) + ½ e^x² · 2x = x e^x²(x² − 1) + x e^x² = x e^x² [x² − 1 + 1] = x³ e^x²，这与原被积函数完全一致。

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8. Common Student Mistakes and How to Avoid Them | 学生常见错误及其避免方法

One typical slip is applying parts directly to x³ e^x² ide becomes confused. Always check that differentiating the chosen v simplifies the remaining integral. Also, forgetting to adjust the constant after back‑substitution costs marks. Practise with similar integrals like ∫ x⁵ e^x² dx or ∫ x e^√x dx to build intuition.

一个典型的失误是不经代换就直接对 x³ e^x² 使用分部积分，导致繁琐的多次分部并出错。另一个错误是在分部积分中误设 dw：如果你把 w 设成 u，微分那一侧就会混乱。务必检查对所选 v 的微分能否简化剩余积分。此外，回代后忘记调整常数也会失分。通过练习相似题目，如 ∫ x⁵ e^x² dx 或 ∫ x e^√x dx，来建立直觉。

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9. Linking to the Edexcel Pure Mathematics Syllabus | 关联 Edexcel 纯数考纲

This hybrid technique appears in Edexcel Units P3 and P4, under 'Integration using substitution and integration by parts'. Questions often combine a polynomial power, an exponential or trigonometric function, and a clever u‑substitution. The mark scheme rewards recognising when a substitution transforms the integral into a standard parts configuration. Familiarise yourself with the general pattern ∫ x²ⁿ⁺¹ e^x² dx, which always reduces to a parts integral after u=x².

这种混合技巧出现在 Edexcel P3 与 P4 单元中，属于'利用代换和分部积分法进行积分'的范畴。题目常将多项式幂、指数函数或三角函数与巧妙的 u-代换结合在一起。评分标准会奖励那些能识别代换可将积分转化为标准分部结构的学生。请熟悉一般模式 ∫ x²ⁿ⁺¹ e^x² dx，它在令 u=x² 之后总能化成一个分部积分。

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10. Extension: Solving Similar Hybrid Integrals | 拓展：求解类似的混合积分

Try these variations to cement your understanding:

• ∫ x e^3x² dx – quick substitution, no parts needed.
• ∫ x³ sin(x²) dx – same substitution, parts on u sin u.
• ∫ x⁵ e^x² dx – substitution then parts on u² e^u, requiring one extra parts cycle.

Each of these follows the PDFjoiner (4)-107 structure and builds fluency with the substitution-first strategy. Practise until you can choose u and set up parts without hesitation.

试试下面的变式以巩固理解：

• ∫ x e^3x² dx —— 简单代换，无需分部。
• ∫ x³ sin(x²) dx —— 同样代换，再对 u sin u 分部。
• ∫ x⁵ e^x² dx —— 代换后再对 u² e^u 分部，需多一次分部过程。

以上各题均遵循 PDFjoiner (4)-107 的结构，能训练你先用代换的策略。反复练习，直到你能毫不犹豫地选择 u 并搭建分部框架。

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11. Strategy Summary Table | 策略总结表

Step	Action	Why
1	Set u = x², du = 2x dx	Eliminates one x, reveals structure
2	Rewrite as (1/2) ∫ u eu du	Prepares for standard parts
3	Parts: v=u, dw=eu du	Differentiating u simplifies
4	Obtain u eu − eu + C₁	Direct integration
5	Multiply by ½ and back-substitute	Final answer in terms of x

步骤 | 操作 | 原因

此表总结了从 PDFjoiner (4)-107 中提炼的解题路线，可作为其他混合积分题的参考模板。

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12. Final Thoughts and Exam Advice | 最后思考与应考建议

Problems like PDFjoiner (4)-107 are designed to test your flexibility. In the exam, scan the integrand for composite functions whose derivatives hide among the other factors. If a direct approach feels forced, pause and consider a substitution. After substituting, the integral should look familiar – if it doesn't, revisit your choice. With steady practice, you'll turn these hybrid integrals from stumbling blocks into scoring opportunities.

像 PDFjoiner (4)-107 这样的题目旨在考察你的灵活性。在考试中，先扫视被积函数，寻找其导数隐藏在其他因子中的复合函数。如果直接硬做感到很牵强，就停下来考虑代换。代换之后，积分应当变得眼熟——如果不是，重新审视你的选择。通过持续练习，你会将这些混合积分从丢分的绊脚石变为得分利器。

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