📚 Exercise 27B: Integration by Substitution | 练习27B：换元积分法
Integration by substitution, often called u-substitution, reverses the chain rule and is essential for handling composite functions in IB Mathematics: Analysis and Approaches. Exercise 27B is designed to build fluency with this method, asking you to spot the inner function and transform integrals systematically. This article unpacks the theory, walks through varied examples, and addresses common pitfalls, helping you move confidently from recognising the pattern to writing the final antiderivative.
换元积分法（也称 u-代换法）是链式法则的逆运算，专门用来处理复合函数的积分，在IB数学分析与方法的课程中举足轻重。练习27B的设计意图就是让你熟练掌握这一技巧，学会识别内层函数并有条理地转换积分。本文会梳理相关理论，讲解不同类型的例子，并分析常见误区，帮助你自信地从辨认结构走向写出最终的原函数。
1. The Reverse of the Chain Rule | 链式法则的逆用
Differentiation tells us that if y = F(g(x)), then y' = F'(g(x))·g' Integration by substitution works backwards: when the integrand has the pattern f(g(x))·g' tructure, choose the right substitution, and rewrite the integral in a simpler variable u.
微分学告诉我们，若 y = F(g(x))，则由链式法则可得 y' = F'(g(x))·g'(x)。换元积分正好反过来：当被积函数具有 f(g(x))·g'(x) 的结构时，只要我们能找到 f 的原函数，就可以还原出 F(g(x))。练习27B的难点就在于识别这种结构，选择合适的换元，并将积分用更简单的变量 u 重新表达。
∫ f(g(x))·g'(x) dx = ∫ f(u) du, where u = g(x)
∫ f(g(x))·g'(x) dx = ∫ f(u) du，其中 u = g(x)
2. The Formal Substitution Procedure | 换元的正式步骤
root, or in a trigonometric, exponential, or logarithmic argument. Next, compute du/dx = g'(x) and rearrange to du = g' hen replace all x-expressions with u equivalents: the original function becomes f(u) and g' grate with respect to u, and finally substitute back x to return to the original variable.
为了避免出错，请遵循一套清晰的步骤。首先，令 u = g(x)，它通常是括号内、根号下、三角函数、指数或对数里面的函数。然后求出 du/dx = g'(x)，并改写成 du = g'(x) dx。接着把所有含 x 的表达式都替换成 u：原被积函数变为 f(u)，而 g'(x)dx 正好变成 du。对 u 积分之后，再把 u 换回 x 的表达式即可。
For definite integrals, you must also change the limits from x-values to u-values, which avoids the need for back-substitution at the end. Exercise 27B includes both indefinite and definite integrals to test this skill.
对于定积分，必须同时把积分上下限从 x 值换成对应的 u 值，这样就不必在最后再换回原变量。练习27B中同时包含不定积分和定积分，就是为了检验对这一细节的掌握。
3. Selecting the Right u | 选择合适的 u
ntegrand. Common candidates include: the inner function of a composite (e.g. u = x³+1 when √x³+1 appears), the exponent in an exponential, the argument of a trigonometric function, or the denominator in a rational expression. If one part of the integrand is clearly the derivative of another, that is often the best u.
换元的艺术在于选对 u，使得变换后的积分变得更简单。你要找导数也出现在被积函数中的那个函数。常见的候选对象包括：复合函数的内部（例如遇到 √x³+1 时可设 u = x³+1）、指数函数的指数部分、三角函数的自变量、有理式的分母等。如果被积函数中某一部分恰好是另一部分的导数，那么它通常就是最佳的 u。
Traditional advice like 'let u be whatever is inside the brackets' works for simple polynomials, but Exercise 27B introduces cases where u is a trigonometric or logarithmic expression, requiring careful inspection.
像“令 u 等于括号里的内容”这样的传统建议，对简单多项式很管用，但练习27B会引入三角函数或对数函数充当 u 的情况，这就需要更仔细的观察。
4. Step-by-Step: Indefinite Integral Example | 逐步演示：不定积分示例
Consider ∫ 3x²√x³+1 dx. Let u = x³+1, then du/dx = 3x², so du = 3x² dx. The integral becomes ∫ √u ting back, the final answer is (2/3)(x³+1)³/² + C. The derivative check confirms correctness: differentiate to get √x³+1·3x².
考虑积分 ∫ 3x²√x³+1 dx。令 u = x³+1，则 du/dx = 3x²，于是 du = 3x² dx。原积分变为 ∫ √u du = ∫ u¹/² du。积分得 (u³/²)/(3/2) + C = (2/3)u³/² + C。换回 x 后，最终答案为 (2/3)(x³+1)³/² + C。求导检验：对结果求导恰好得到 √x³+1·3x²，验证无误。
Always verify your antiderivative by mentally differentiating it. This helps you catch algebraic slips in the substitution and integration stages.
养成在脑中求导检验原函数的习惯，这能帮你发现在换元或积分过程中出现的代数疏漏。
5. Trigonometric Substitutions: Sine and Cosine | 三角换元：正弦与余弦
omes (1/2) ∫ u du = (1/4) u² + C = (1/4) sin²(2x) + C. Alternatively, u = cos(2x) leads to –(1/4) cos²(2x) + C – seemingly different but equivalent via the identity sin²θ + cos²θ = 1.
三角函数的积分常需要意识到正弦与余弦的导数互为关联。对于 ∫ sin(2x) cos(2x) dx，可设 u = sin(2x)，则 du = 2 cos(2x) dx，从而 (1/2) du = cos(2x) dx。积分变为 (1/2) ∫ u du = (1/4) u² + C = (1/4) sin²(2x) + C。若改用 u = cos(2x)，会得到 –(1/4) cos²(2x) + C，两者看似不同，但利用恒等式 sin²θ + cos²θ = 1 可证它们等价。
Exercise 27B may present integrals like ∫ tan x dx = –ln |cos x| atch for these quotients; the numerator often resembles the derivative of the denominator.
练习27B中可能会出现 ∫ tan x dx = –ln |cos x| + C 这样的积分，解法是把 tan x 写成 sin x/cos x 并令 u = cos x。注意这类商的模式：分子常常是分母的导数。
6. Exponential and Logarithmic Integrands | 指数与对数被积函数
For ∫ esin x natural choice is u = sin x, since du = cos x dx. The integral simplifies directly to ∫ eu du = eu + C = esin x (ln x)² + C. These examples show that substitution can reduce exponential and logarithmic expressions to standard forms.
对于 ∫ esin x cos x dx，很自然的选择是 u = sin x，因为 du = cos x dx。积分直接简化为 ∫ eu du = eu + C = esin x + C。而在 ∫ (ln x)/x dx 中，令 u = ln x，du = (1/x) dx，则积分变为 ∫ u du = (1/2) u² + C = (1/2)(ln x)² + C。这些例子表明，换元可以把指数和对数表达式化归为标准形式。
a fraction and the numerator is exactly the derivative of the denominator, leading to ∫ (g'(x)/g(x)) dx = ln |g(x)| C. Recognising this directly saves the formal substitution when it is not strictly needed.
还要能识别一类特殊积分：分子恰好是分母的导数，此时可直接写出 ∫ (g'(x)/g(x)) dx = ln |g(x)| + C。当换元并非严格必要时，直接识别这一模式能节省时间。
7. Handling Definite Integrals with Substitution | 处理需换元的定积分
du = 2x dx. When x = 0, u = 1; when x = 1, u = 2. The integral becomes ∫₁² (1/u) du = [ln |u|]₁² = ln 2 – n 2. There is no need to return to x, which simplifies the calculation and reduces errors.
对定积分进行换元时，必须同时转换被积函数和积分限。例如 ∫₀¹ 2x/(x²+1) dx，令 u = x²+1，则 du = 2x dx。当 x = 0 时 u = 1；当 x = 1 时 u = 2。积分变为 ∫₁² (1/u) du = [ln |u|]₁² = ln 2 – ln 1 = ln 2。无需再换回 x，这简化了计算也降低了出错率。
Always write the new limits clearly next to the substitution step. A common mistake is leaving the original x-limits on the u-integral, which produces a meaningless numerical result.
务必在换元步骤旁清晰写出新的上下限。常见错误之一就是 u 积分上还保留着原有的 x 对应限值，导致数值结果失去意义。
8. Hidden Substitutions and Algebraic Twists | 隐藏的换元与代数技巧
Some integrals in Exercise 27B disguise the substitution. For example, ∫ x√x+1 dx appears not to have a derivative-du pair. Let u = x+1, so x = u – 1 and dx = du. The integral becomes ∫ (u–1)√u du = ∫ (u³/² – u¹/²) du, which is easily integrated. This technique – substituting the binomial inside a root – is a valuable extension of the basic idea.
练习27B中有些积分会隐藏换元。例如 ∫ x√x+1 dx 看起来并不具备“导数-du”的配对。令 u = x+1，则 x = u – 1，dx = du。积分转化为 ∫ (u–1)√u du = ∫ (u³/² – u¹/²) du，轻松积分。这种将根号内的二项式整个替换的方法，是基础换元的有效拓展。
Another twist: ∫ (1/(1+ex)) dx. Multiply numerator and denominator by e–x to get ∫ e–x/(e–x+1) dx, then let u = e–x+1. Such algebraic manipulation before substitution is a sharp test of insight and is typical of harder IB problems.
另一种变形：∫ (1/(1+ex)) dx。分子分母同乘以 e–x 后得到 ∫ e–x/(e–x+1) dx，再令 u = e–x+1。这种在换元前进行的代数调整，是对洞察力的严格考验，也是IB较难题目中的常见手法。
9. Common Errors in Substitution | 换元中的常见错误
takes cluster around the derivative factor. Substituting u but forgetting to account for g' error. Others include: misplacing the differential (e.g. writing du = dx instead of du = g' stant adjustment, e.g. du = 2x dx but the integral has only x dx, requiring a factor of ½.
最常见的错误集中在导数因子上。选对了 u 却忘记处理 g'(x)，是出现频率最高的失误。其他常见问题还包括：错误地写出 du = dx 而非 du = g'(x) dx、把和式的积分当作乘积去换元、定积分忘记改上下限。还有一个陷阱是导数可能需要常数调整，例如 du = 2x dx 但被积函数中只有 x dx，此时需要引入因子 ½。
- Forgetting to substitute back to x: Leaving the answer in terms of u is marked as incomplete.
- Using same variable name in substitution: Avoid notation like x = x²+1; use a new variable.
- Misplacing constants: Always separate multiplicative constants outside the integral.
- 忘记换回 x：答案保留 u 的形式会被视为未完成。
- 换元时变量名混用：避免使用类似 x = x²+1 的记法；一定要引入新变量。
- 常数处理不当：永远先把常数因子提到积分号外面。
10. Practice Strategies for Exercise 27B | 练习27B的解题策略
Start each integral by scanning for a function and its derivative. Ask: 'If I let u be this expression, does du appear in the integrand?' Next, identify any missing constant and adjust with a factor outside. When stuck, try a substitution that simplifies the most complicated part, even if the derivative is not immediately obvious. With definite integrals, always draw a small table mapping x-limits to u-limits to keep the process organised.
面对每一个积分，先扫视是否存在一个函数及其导数。自问：“如果让 u 等于这个表达式， du 会出现在被积函数中吗？”接着，识别出任何缺失的常数，并用外部因子加以调整。遇到困难时，不妨尝试用一个能简化最复杂部分的换元，哪怕它的导数并不一目了然。对于定积分，一定要画一个简表，将 x 的上下限映射到对应的 u 值，让过程井井有条。
Work through Exercise 27B by categorising problems: polynomials inside roots, trigonometric products, exponential compositions, and rational functions. After completing a batch, differentiate your answers to check them mentally – this builds speed and confidence for the exam.
做练习27B时，可以把题目分类：根号内的多项式、三角函数的乘积、指数复合函数、以及有理函数。每完成一组，就在脑海中给答案求导检验——这样的习惯能提升考试中的速度与信心。
| Integrand | Substitution | Result |
| ∫ (3x²+2x) ex³+x² dx | u = x³+x² | ex³+x² + C |
| ∫ cos x / (1+sin x) dx | u = 1+sin x | ln |1+sin x| + C |
| ∫ x² √x³–5 dx | u = x³–5 | (2/9)(x³–5)³/² + C |
| ∫₁e (ln x)²/(x) dx | u = ln x | 1/3 |
11. Substitution and the Chain Rule: A Unified View | 换元与链式法则的统一视角
s. If you differentiate (2/3)(x³+1)³/², the chain rule yields (3/2)(2/3) (x³+1)¹/² · 3x² = √x³+1·3x². Substitution simply reconstructs this process backwards. This conceptual link also helps in differential equations, where separating variables uses a similar logic of differential manipulation.
把换元视为链式法则的逆微分，能让你明白为什么这种方法管用。若对 (2/3)(x³+1)³/² 求导，链式法则会给出 (3/2)(2/3) (x³+1)¹/² · 3x² = √x³+1·3x²。换元不过是将这一过程逆向重演。这一概念联系在微分方程中同样有用，分离变量法就用到了类似的微分操作逻辑。
12. Summary and Final Tips | 总结与考前贴士
Integration by substitution transforms a complicated integral into a basic one by changing the variable. The key is pattern recognition: spot a function and its derivative. Remember to adjust constants, change limits for definite integrals, and always check by differentiating your result. Exercise 27B is carefully constructed to build these reflexes; treat it as a diagnostic of your integration fluency.
换元积分通过变量代换将复杂积分转化为基本积分。关键在于模式识别：发现一个函数及其导数。牢记调整常数、定积分改变上下限、并始终以求导检验结果。练习27B被精心设计来培养这些反应能力，请把它当作检验你积分熟练度的诊断工具。
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