Exercise 15H.1 | 练习15H.1

📚 Exercise 15H.1 | 练习15H.1

Exercise 15H.1 in your IB Mathematics course is a carefully curated set of problems designed to build fluency in one of the most powerful integration techniques: integration by substitution. Often the first advanced method students meet after mastering basic antiderivatives, it transforms a complicated integral into a standard form by introducing a new variable. This article walks through the key concepts behind substitution, provides step-by-step model solutions that mirror the style of textbook exercises, and highlights common pitfalls to help you approach every integral in this set with confidence.

IB数学课程中的练习15H.1是一组精心编排的题目，旨在帮助你熟练掌握一种极为强大的积分技巧：换元积分法。它通常是你掌握基本原函数后遇到的第一个进阶方法，通过引入新变量将复杂的积分转化为标准形式。本文将梳理换元法的核心概念，提供与课本练习风格一致的逐步示范解答，并指出常见陷阱，让你能够自信地应对该练习中的每一道积分题。

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1. What Is Integration by Substitution? | 什么是换元积分法？

Integration by substitution is the reverse process of the chain rule for differentiation. If you recognise that an integrand contains a function and its derivative – or can be manipulated to contain them – you can let u equal the inner function, rewrite the integral entirely in terms of u, integrate simply, and then substitute back. This method is sometimes called u-substitution.

换元积分法是微分链式法则的逆过程。如果你发现被积函数包含一个函数及其导数——或者可以变形为包含它们——就可以令 u 等于内层函数，将整个积分用 u 重新表达，轻松积分后再换回原变量。这种方法也常被称作 u-代换。

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2. The Chain Rule in Reverse | 逆向链式法则

Recall the chain rule: d/dx [f(g(x))] = f'(g(x)) · g'(x). Integrating both sides gives ∫ f'(g(x)) · g'(x) dx = f(g(x)) + C. Substitution formalises this: set u = g(x), so du/dx = g'(x) ⇒ du = g'(x) dx. The integral becomes ∫ f'(u) du = f(u) + C. In Exercise 15H.1, nearly every question can be deconstructed using this logic.

回想链式法则：d/dx [f(g(x))] = f'(g(x)) · g'(x)。两边积分得到 ∫ f'(g(x)) · g'(x) dx = f(g(x)) + C。换元法将这个过程规范化：设 u = g(x)，则 du/dx = g'(x) ⇒ du = g'(x) dx。积分变为 ∫ f'(u) du = f(u) + C。在练习15H.1中，几乎每一道题都可以用这个逻辑拆解。

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3. Choosing u: The LIATE Guideline | 选择 u：LIATE 法则

A common question is which part of the integrand to set as u. While substitution problems in Exercise 15H.1 are designed to make the choice obvious, a useful heuristic is LIATE: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Choose u as the function type that appears first in this list. For example, in ∫ x · e^(x²) dx, the algebraic x is a good candidate because its derivative 1 is simple, but here the inner function x² appears in the exponent, so u = x² works perfectly.

常见的问题是应该把被积函数中的哪一部分设为 u。虽然练习15H.1中的换元题目在设计上已经让选择变得明显，但一个有用的启发式规则是 LIATE：对数、反三角、代数、三角、指数。选择在列表中先出现的函数类型作为 u。例如在 ∫ x · e^(x²) dx 中，代数部分 x 的导数很简单，但这里指数部分的内部函数是 x²，因此令 u = x² 最为理想。

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4. Example 1: Basic Polynomial Substitution | 示例1：基本多项式换元

Consider ∫ 2x (x² + 1)³ dx. Let u = x² + 1. Then du/dx = 2x, so du = 2x dx. The integral becomes ∫ u³ du = (1/4) u⁴ + C. Substitute back: (1/4)(x² + 1)⁴ + C. This pattern appears repeatedly in the early questions of Exercise 15H.1.

考虑 ∫ 2x (x² + 1)³ dx。令 u = x² + 1，则 du/dx = 2x，所以 du = 2x dx。积分变为 ∫ u³ du = (1/4) u⁴ + C。代回原变量：(1/4)(x² + 1)⁴ + C。这种模式在练习15H.1的开头几题中反复出现。

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5. Example 2: Substitution with a Root | 示例2：包含根号的换元

Evaluate ∫ x √(2x² + 3) dx. Spot the inner function 2x² + 3, whose derivative is 4x. We have x dx, which is a constant multiple of 4x dx. Set u = 2x² + 3, then du = 4x dx ⇒ (1/4) du = x dx. Rewrite: ∫ √u · (1/4) du = (1/4) ∫ u^(1/2) du = (1/4) · (2/3) u^(3/2) + C = (1/6)(2x² + 3)^(3/2) + C.

计算 ∫ x √(2x² + 3) dx。发现内层函数 2x² + 3 的导数是 4x。被积函数中有 x dx，正好是 4x dx 的常数倍。设 u = 2x² + 3，则 du = 4x dx ⇒ (1/4) du = x dx。改写积分：∫ √u · (1/4) du = (1/4) ∫ u^(1/2) du = (1/4) · (2/3) u^(3/2) + C = (1/6)(2x² + 3)^(3/2) + C。

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6. Example 3: Trigonometric Substitution | 示例3：三角换元

For ∫ sin³ x cos x dx, letting u = sin x is natural because du = cos x dx. Then the integral becomes ∫ u³ du = (1/4) sin⁴ x + C. Always check that the substitution covers the entire integrand; leftover x terms usually signal the wrong choice of u.

对于 ∫ sin³ x cos x dx，很自然会令 u = sin x，因为 du = cos x dx。这样积分变为 ∫ u³ du = (1/4) sin⁴ x + C。务必检查代换是否覆盖了整个被积函数；如果换完后还残留原变量项，通常说明 u 的选择有问题。

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7. Example 4: Exponential Functions | 示例4：指数函数换元

Even when e^x appears, substitution is often needed. Try ∫ e^(2x) √(1 + e^(2x)) dx. The inner function (1 + e^(2x)) has derivative 2e^(2x), which matches the e^(2x) dx part up to a constant. Let u = 1 + e^(2x), du = 2e^(2x) dx ⇒ (1/2) du = e^(2x) dx. The integral is ∫ √u · (1/2) du = (1/2) · (2/3) u^(3/2) + C = (1/3)(1 + e^(2x))^(3/2) + C. Such problems appear towards the end of Exercise 15H.1 to test flexibility.

即使出现 e^x，换元法也常常必不可少。尝试 ∫ e^(2x) √(1 + e^(2x)) dx。内层函数 (1 + e^(2x)) 的导数是 2e^(2x)，正好与被积函数中的 e^(2x) dx 仅差一个常数倍。令 u = 1 + e^(2x)，du = 2e^(2x) dx ⇒ (1/2) du = e^(2x) dx。积分变为 ∫ √u · (1/2) du = (1/2) · (2/3) u^(3/2) + C = (1/3)(1 + e^(2x))^(3/2) + C。这类题目通常出现在练习15H.1的后段，用来检验你的灵活应用能力。

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8. Handling Definite Integrals with Substitution | 定积分的换元处理

When the integral has limits, you have two options: either change the limits to match the u-variable, or integrate in u and then revert to x before applying the original limits. The first method is cleaner. For ∫ from 0 to 1 of 2x (x²+1)³ dx, with u = x²+1: when x=0, u=1; when x=1, u=2. The integral becomes ∫ from 1 to 2 of u³ du = [u⁴/4] from 1 to 2 = 16/4 − 1/4 = 15/4. Never forget to update limits or you will get the wrong numerical answer.

当积分带有上下限时，你有两种处理方式：要么将上下限转换为与 u 变量一致的数值，要么先在 u 下积分，换回 x 后再代入原上下限。第一种方法更简洁。对于 ∫ ₀¹ 2x (x²+1)³ dx，令 u = x²+1：当 x=0 时 u=1；当 x=1 时 u=2。积分变为 ∫ ₁² u³ du = [u⁴/4]₁² = 16/4 − 1/4 = 15/4。切勿忘记更新积分限，否则会得到错误的数值答案。

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9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

The most frequent error in Exercise 15H.1 is forgetting to replace dx entirely. Always compute du = (du/dx) dx and isolate dx. If your new integral still contains x, the substitution is incomplete. Another mistake is misaligning the constant multiple. If du = 4x dx but you have x dx, extract (1/4) carefully. Finally, when using definite integrals, failing to change limits leads to a classic “mixed variable” error. Pause after substituting limits to verify consistency.

在练习15H.1中最常见的错误是忘记把 dx 完全替换掉。务必计算 du = (du/dx) dx 并解出 dx。如果新积分中仍然含有 x，就说明代换不彻底。另一个错误是常数倍处理不当。如果 du = 4x dx 但你只有 x dx，一定要小心提取出 (1/4)。最后，对于定积分，忘记更换积分限会导致经典的“变量混用”错误。替换完积分限后不妨暂停一下，检查是否统一。

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10. Tips for Spotting Substitution Quickly | 快速识别换元法的技巧

Scan the integrand for two factors: one factor that looks like a composition f(g(x)), and another that looks like a multiple of g'(x). In Exercise 15H.1, the inner function is often raised to a power, inside a square root, or in a denominator. Train your eye to see “a function and its derivative almost multiplied together”. With practice, you will recognise patterns like ∫ x/√(1−x²) dx where u = 1−x² works smoothly.

快速扫视被积函数，寻找两种因子：一个是像复合函数 f(g(x)) 的部分，另一个是像 g'(x) 的常数倍。在练习15H.1中，内层函数经常以幂次形式出现，或在根号内、分母中。训练自己的眼光去识别“一个函数与它的导数几乎相乘”的结构。通过练习，你会迅速认出像 ∫ x/√(1−x²) dx 这样的模式，令 u = 1−x² 就能顺利解决。

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11. Connecting Substitution to Other Integration Methods | 换元法与其他积分技巧的关联

Substitution is not an isolated technique. It often serves as a first step before integration by parts or partial fractions. For instance, an expression like ∫ ln x / x dx is handled by u = ln x, reducing to a simple power integral. In later chapters, you will combine substitution with trigonometric identities to tackle integrals of the form ∫ sin²x dx. Mastering Exercise 15H.1 therefore lays the groundwork for the entire integration toolkit.

换元法并不是一个孤立的技巧。它常常作为分部积分或有理分式分解前的第一步。例如，∫ ln x / x dx 可令 u = ln x 来处理，化归为一个简单的幂函数积分。在后续章节中，你需要将换元法与三角恒等式结合，以求解像 ∫ sin²x dx 这类积分。因此，掌握练习15H.1为你的整个积分工具箱奠定了坚实基础。

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12. Summary and Next Steps | 总结与下一步

Exercise 15H.1 provides a structured path to internalise u-substitution. Begin by verifying your choice of u with the derivative check, rewrite the integral cleanly, integrate, and substitute back. For definite integrals, update limits straight away. Review each completed problem not only for the answer but to understand why the substitution worked; this reflective practice will make future exercises feel intuitive. Once you finish, try the mixed review exercises to test your ability to choose methods independently.

练习15H.1提供了一条结构化的路径，帮助你内化 u-代换法。从用导数检验 u 的选择开始，清晰地改写积分，求出原函数，再换回原变量；对于定积分，立即更新积分限。做完每道题后，不仅要核对答案，更要反思为什么这个代换有效；这种复盘能让后续的练习变得直觉化。完成这一组练习后，不妨尝试综合复习题，检验自己独立选择方法的能力。

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