📚 Exercise 24C.2: Integration by Substitution | 练习24C.2：换元积分法
The ability to evaluate integrals using substitution, often called u-substitution, is a cornerstone of IB Mathematics: Analysis and Approaches. Exercise 24C.2 is designed to strengthen your skill in recognising when and how to apply this method, covering a range of functions from simple powers to trigonometric and exponential combinations. In this article, we will walk through the key principles, worked examples that mirror typical IB problems, and strategies that will enable you to complete Exercise 24C.2 with precision and confidence.
运用代换积分法（通常称为u代换）计算积分的能力是IB数学分析与方法课程的核心基础。练习24C.2通过从简单幂函数到三角函数和指数复合函数的一系列题目，旨在强化你对何时以及如何运用这一方法的识别能力与计算技巧。本文将带你梳理核心原理，通过具有IB风格的典型例题讲解和解题策略，帮助你精准、自信地完成练习24C.2。
1. Understanding the Chain Rule in Reverse | 逆向领悟链式法则
Integration by substitution is essentially the reverse of the chain rule for differentiation. When we differentiate f(g(x)), the chain rule gives f'(g(x)) · g'(x). Therefore, if an integral contains a composite function alongside the derivative of its inner function up to a constant factor, substitution can unravel it. Recognising this structure is the first critical step.
换元积分法本质上是链式法则的逆向操作。当我们对f(g(x))求导时，链式法则给出f'(g(x)) · g'(x)。因此，如果一个积分中含有复合函数，并且被积函数中还伴有该复合函数内层函数的导数（至多相差一个常数因子），代换技巧就能将其解开。识别这种结构是至关重要的第一步。
2. The Substitution Method: Step-by-Step | 代换积分法：逐步操作
procedure follows these stages: (i) Choose a suitable u = g(x) whose derivative du/dx = g' (ii) Differentiate to find du/dx and rearrange to express dx in terms of du, e.g., du = g' v) Finally, replace u by g(x) to return the answer in terms of the original variable. For definite integrals, the limits must also be changed to u-values.
标准的u代换步骤如下：(i) 选取恰当的u = g(x)，使得它的导数du/dx = g'(x)出现在被积函数中。(ii) 求导以得到du/dx，通过移项用du表示dx，例如du = g'(x) dx。(iii) 把所有含x的表达式和dx都替换掉，将积分完全转换为关于u的积分。(iv) 计算新的关于u的积分。(v) 最后用g(x)回代u，将答案重新表示为原变量的函数。对于定积分，积分限也必须转换为u值。
3. Choosing the Right Substitution | 选择正确的代换
A common challenge is deciding what to set as u. General guidelines include: let u be the inner function of a composite, especially when its derivative appears elsewhere; let u be the denominator in a rational integrand if the numerator is proportional to its derivative; let u be the argument of a trigonometric, exponential or logarithmic function when multiplied by a derivative-like factor. Practice and familiarity with derivative families build intuition.
最常见的一个难题是如何决定该设什么为u。一般准则包括：选取复合函数的内层函数作为u，特别是当它的导数也出现在别处时；在被积函数为有理式时，若分子与分母的导数成比例，可取分母为u；当三角函数、指数或对数函数的自变量与一个形如导数的因子相乘时，可取该自变量为u。通过大量练习熟悉各类导数函数族，将逐步建立起直觉。
4. Example 1: Recognising f '(x) · [f(x)]ⁿ | 示例1：识别 f '(x)·[f(x)]ⁿ 模式
Consider ∫ (3x² + 2x) √x³ + x² tches the factor outside. Let u = x³ + x². Then du = (3x² + 2x) dx. The integral becomes ∫ √u du = ∫ u1/2 du = (2/3) u3/2 + C. Substituting back gives (2/3)(x³ + x²)3/2 + C.
假设积分 ∫ (3x² + 2x) √x³ + x² dx。注意到被开方数x³ + x²的导数恰好是外部因式3x² + 2x。设u = x³ + x²，则du = (3x² + 2x) dx。原积分化为 ∫ √u du = ∫ u1/2 du = (2/3) u3/2 + C。回代后即得 (2/3)(x³ + x²)3/2 + C。
5. Example 2: Trigonometric Substitution for Powers | 示例2：三角函数的幂次代换
illustrates how isolating one factor as the derivative of the other simplifies trigonometric integrals rapidly.
计算积分 ∫ sin⁵ x cos x dx。sin x的导数是cos x，因此设u = sin x，则du = cos x dx。积分化为 ∫ u⁵ du = (1/6) u⁶ + C = (1/6) sin⁶ x + C。这个例子表明，将其中一个因式视为另一部分的导数能迅速简化三角函数的积分。
6. Example 3: Rational Function Integration | 示例3：有理函数积分
e numerator. Let u = x² + x + 1, then du = (2x+1) dx. The integral becomes ∫ (1/u) du = ln|u| + C = ln|x² + x + 1| + C. This pattern is extremely common in IB exercises.
计算 ∫ (2x+1) / (x² + x + 1) dx。分母的导数是2x+1，正好是分子。设u = x² + x + 1，得du = (2x+1) dx。积分化为 ∫ (1/u) du = ln|u| + C = ln|x² + x + 1| + C。这种模式在IB练习中极为常见。
7. Example 4: Substitution with Exponential Expressions | 示例4：指数表达式的代换
Solve ∫ x ex² the exponential. Set u = x², then du = 2x dx, so x dx = (1/2) du. The integral becomes ∫ eu · (1/2) du = (1/2) eu + C = (1/2) ex² + C. Remember to adjust for constant factors introduced by du.
求解 ∫ x ex² dx。x²的导数是2x，与指数函数外的x仅相差常数倍。设u = x²，则du = 2x dx，因此x dx = (1/2) du。积分化为 ∫ eu · (1/2) du = (1/2) eu + C = (1/2) ex² + C。要注意根据du调整常数因子。
8. Definite Integrals and Changing Limits | 定积分与积分限的转换
For definite integrals like ∫₀¹ 3x² √x³ + 1 u = x³ + 1, so du = 3x² dx. When x = 0, u = 1; when x = 1, u = 2. The integral becomes ∫₁² √u du = [ (2/3) u3/2 ]₁² = (2/3)(2√2 - 1). Avoid substituting back to x-values by directly evaluating with the new u-limits.
对于像 ∫₀¹ 3x² √x³ + 1 dx 这样的定积分，只要转换积分限，代换法同样清晰易行。设u = x³ + 1，得du = 3x² dx。当x = 0时，u = 1；当x = 1时，u = 2。积分化为 ∫₁² √u du = [ (2/3) u3/2 ]₁² = (2/3)(2√2 - 1)。直接使用新的u限进行计算，无需再回代为x值。
9. Common Pitfalls and How to Avoid Them | 常见陷阱与防范
Mistakes often arise from incomplete substitution or forgetting to replace dx with du/g'(x). Always express the entire integrand in terms of u, including all x' ou must introduce a factor of 1/2. Also, with definite integrals, never forget to update the limits or you may obtain nonsensical results. Constant factors should be carefully carried through each step.
错误通常源于代换不彻底或忘记用du/g'(x)替换dx。一定要将整个被积函数都用u表示，包括所有x项。另一个易犯错误是没能正确识别du因子；例如在∫ sin(2x) dx中，设u = 2x可得du = 2 dx，此时必须引入因子1/2。对于定积分，切不可忘记更新积分限，否则可能得出荒谬的结果。每一个常数因子都应谨慎地随计算步骤传递。
10. Practice Strategies for Exercise 24C.2 | 练习24C.2的实战策略
function, differentiate it, and see if the remaining part of the integrand matches du, possibly up to a constant. If not, consider alternative u-choices or splitting the integrand. Practice both indefinite and definite forms, and check your answers by differentiation. Group similar problems and reflect on why the same substitution works. Consistent practice with the methods demonstrated here will make substitution intuitive and automatic.
要在练习24C.2中取得成功，首先扫视被积函数，寻找一个函数与其导数。尝试将内层函数设为u，对其求导，检查被积函数剩余部分是否与du匹配，或许仅差常数倍。如果不匹配，考虑其他u的选择或者拆分被积函数。同时练习不定积分和定积分形式，并通过求导检验答案。将同类问题归类，反思为什么同样的代换方式有效。反复运用本文所展示的方法进行训练，代换将变得直观而自动。
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