Mastering Exercise 21J.1: Integration by Substitution | 攻克练习21J.1：换元积分法

📚 Mastering Exercise 21J.1: Integration by Substitution | 攻克练习21J.1：换元积分法

Exercise 21J.1 is a classic set of integration problems found in many IB Mathematics textbooks, focusing entirely on the method of u-substitution. This technique, also called integration by substitution, is the reverse of the chain rule and is essential for handling integrals where an inner function and its derivative appear together. In this article, we will walk through the key ideas and solve typical problems from Exercise 21J.1, breaking down each step in both English and Chinese to support bilingual learners. By the end, you will be able to recognise substitution patterns, choose the right u, and transform even messy integrals into simple power or trigonometric forms.

练习21J.1是许多IB数学教材中的经典积分题组，完全围绕换元积分法（u-代换）展开。这一技巧也被称为“代入积分法”，是链式法则的逆运算，对于处理被积函数中同时出现内层函数及其导数的情况至关重要。本文中，我们将梳理核心要点并解答练习21J.1中的典型题目，用中英双语逐步拆解每一个步骤。学完本文，你将能够识别代换模式、正确选择 u，并将看似复杂的积分转化为简单的幂函数或三角形式。

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1. The Core Idea of u-Substitution | u-代换的核心思想

At the heart of Exercise 21J.1 lies a single concept: if an integrand can be written as f'(g(x))·g'(x), then the substitution u = g(x) simplifies the integral to ∫ f'(u) du = f(u) + C. In practice, we look for a function whose derivative also appears as a factor, or can be easily obtained by algebraic manipulation.

练习21J.1的核心思想是：如果被积函数可以写成 f'(g(x))·g'(x) 的形式，那么令 u = g(x) 就能把积分简化为 ∫ f'(u) du = f(u) + C。实际操作中，我们寻找一个函数，它的导数也以因式的形式出现，或者可以通过简单的代数变形得到。

For example, consider ∫ 2x·cos(x²) dx. Here g(x) = x² and g'(x) = 2x. Letting u = x² gives du = 2x dx, and the integral becomes ∫ cos u du = sin u + C = sin(x²) + C.

例如，考虑 ∫ 2x·cos(x²) dx。这里 g(x) = x²，g'(x) = 2x。令 u = x² 得到 du = 2x dx，积分变为 ∫ cos u du = sin u + C = sin(x²) + C。

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2. Choosing u: The Detective Work | 选择 u：像侦探一样思考

Selecting the correct u is the most important skill in Exercise 21J.1. Start by scanning for a composite function – an expression inside parentheses, under a root, in an exponent, or as the argument of a trig function. Then check if its derivative is present elsewhere in the integrand, possibly up to a constant multiple.

选择正确的 u 是练习21J.1中最关键的能力。首先扫视被积函数，寻找复合函数——括号内的表达式、根号下的式子、指数上的量，或是三角函数的自变量。然后检查它的导数是否出现在被积函数的其他部分，即使差一个常数倍数也可以。

If the derivative is missing, do not force the substitution. Some integrals in the exercise may require a little rearrangement, such as multiplying and dividing by a constant, before the substitution becomes visible.

如果导数没有直接出现，不要强行代换。练习中的某些积分可能需要一点变形，比如先乘以再除以某个常数，才能让代换关系显现出来。

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3. Problem 1: Polynomial Raised to a Power | 例题1：多项式的高次幂

A typical first problem in Exercise 21J.1 is ∫ x(x²+1)³ dx. The inner function is clearly x²+1, and its derivative 2x is almost the x factor. We can adjust by writing x dx = (1/2) du.

练习21J.1中一个典型的开篇题目是 ∫ x(x²+1)³ dx。内层函数显然是 x²+1，其导数 2x 与因式 x 非常接近，我们可以通过 x dx = (1/2) du 进行调整。

Set u = x²+1, then du = 2x dx ⇒ x dx = (1/2) du. The integral becomes ∫ u³ · (1/2) du = (1/2)·(u⁴/4) + C = u⁴/8 + C. Substituting back gives (x²+1)⁴/8 + C.

令 u = x²+1，则 du = 2x dx ⇒ x dx = (1/2) du。积分变为 ∫ u³ · (1/2) du = (1/2)·(u⁴/4) + C = u⁴/8 + C。回代得到 (x²+1)⁴/8 + C。

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4. Problem 2: Trigonometric Powers | 例题2：三角函数的幂次

Consider ∫ sin⁴x cos x dx. Here the composite function is sin x raised to the 4th power, and its derivative cos x is sitting right next to it. This is a perfect u-substitution setup.

考虑 ∫ sin⁴x cos x dx。这里的复合函数是 sin x 的四次方，而它的导数 cos x 紧挨在旁边。这是一个完美的 u-代换结构。

Let u = sin x, so du = cos x dx. The integral collapses to ∫ u⁴ du = u⁵/5 + C = (sin⁵x)/5 + C. Exercise 21J.1 reinforces the idea that when you see a trig function and its derivative paired, substitution is almost immediate.

令 u = sin x，则 du = cos x dx。积分收缩为 ∫ u⁴ du = u⁵/5 + C = (sin⁵x)/5 + C。练习21J.1强化了一个概念：当你看到一个三角函数与其导数成对出现时，代换几乎可以瞬间完成。

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5. Problem 3: Exponential Composition | 例题3：指数复合型

Another common style in Exercise 21J.1 is ∫ e^(3x) √(1+e^(3x)) dx. Look inside the square root: the inner function is 1+e^(3x). Its derivative is 3e^(3x), which appears in the integrand. A constant factor of 3 needs to be introduced.

练习21J.1中另一常见类型是 ∫ e^(3x) √(1+e^(3x)) dx。观察根号内：内层函数是 1+e^(3x)。它的导数是 3e^(3x)，在被积函数中出现了 e^(3x)，只需引入常数因子3。

Set u = 1+e^(3x), then du = 3e^(3x) dx, so e^(3x) dx = (1/3) du. The integral becomes ∫ √u · (1/3) du = (1/3) ∫ u^(1/2) du = (1/3)·(2/3)u^(3/2) + C = (2/9)(1+e^(3x))^(3/2) + C.

令 u = 1+e^(3x)，则 du = 3e^(3x) dx，于是 e^(3x) dx = (1/3) du。积分化为 ∫ √u · (1/3) du = (1/3) ∫ u^(1/2) du = (1/3)·(2/3)u^(3/2) + C = (2/9)(1+e^(3x))^(3/2) + C。

∫ e^(3x)√(1+e^(3x)) dx = 2/9 (1+e^(3x))^(3/2) + C

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6. Problem 4: Logarithmic Integrals | 例题4：对数积分

The integral ∫ (ln x)/x dx appears frequently in Exercise 21J.1. The derivative of ln x is 1/x, which multiplies the integrand. This invites the substitution u = ln x.

积分 ∫ (ln x)/x dx 在练习21J.1中频繁出现。ln x 的导数是 1/x，而被积函数正是 ln x 乘以 1/x，这提示我们使用 u = ln x 的代换。

Let u = ln x, then du = (1/x) dx. The integral transforms to ∫ u du = u²/2 + C = (ln x)²/2 + C. This pattern extends to any power of ln x over x, such as ∫ (ln x)⁵/x dx = (ln x)⁶/6 + C.

令 u = ln x，则 du = (1/x) dx。积分转化为 ∫ u du = u²/2 + C = (ln x)²/2 + C。这一模式可以推广到任何 (ln x)ⁿ/x 的形式，例如 ∫ (ln x)⁵/x dx = (ln x)⁶/6 + C。

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7. Problem 5: Rationalising Substitutions | 例题5：有理化代换

Exercise 21J.1 also introduces integrals like ∫ x/√(1−x²) dx. The inner function is 1−x² inside a square root, and its derivative −2x is present up to a constant. This substitution will eliminate the root.

练习21J.1还引入了诸如 ∫ x/√(1−x²) dx 的积分。内层函数 1−x² 位于平方根下，其导数 −2x 以常数倍的形式出现。这一代换将消去根号。

Set u = 1−x², so du = −2x dx ⇒ x dx = −(1/2) du. The integral becomes ∫ 1/√u · (−1/2) du = −(1/2) ∫ u^(−1/2) du = −(1/2)·2u^(1/2) + C = −√(1−x²) + C.

令 u = 1−x²，则 du = −2x dx ⇒ x dx = −(1/2) du。积分化为 ∫ 1/√u · (−1/2) du = −(1/2) ∫ u^(−1/2) du = −(1/2)·2u^(1/2) + C = −√(1−x²) + C。

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8. Problem 6: Recognising Inverse Trigonometric Forms | 例题6：识别反三角函数形式

Some integrals in Exercise 21J.1 are cleverly designed to lead to inverse trig functions after substitution. For instance, ∫ 1/(1+4x²) dx does not immediately suggest a u, but a small rewrite reveals the pattern. Let u = 2x, then du = 2 dx, and the integral becomes (1/2) ∫ 1/(1+u²) du = (1/2) arctan u + C = (1/2) arctan(2x) + C.

练习21J.1中的某些积分设计巧妙，经过代换后会导出反三角函数。例如 ∫ 1/(1+4x²) dx 起初看不出 u，但稍加改写即可显现模式。令 u = 2x，则 du = 2 dx，积分变为 (1/2) ∫ 1/(1+u²) du = (1/2) arctan u + C = (1/2) arctan(2x) + C。

This problem highlights that substitution can reveal standard forms like ∫ 1/(1+u²) du and ∫ 1/√(1−u²) du, which are fundamental in IB Mathematics.

这道题强调代换能够揭示标准形式，如 ∫ 1/(1+u²) du 和 ∫ 1/√(1−u²) du，这些在IB数学中是基础公式。

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9. Common Pitfalls in Exercise 21J.1 | 练习21J.1中的常见陷阱

One frequent mistake is forgetting to adjust for the constant factor when the derivative does not exactly match. Always set du = g'(x) dx and then solve for the needed expression. If the integrand has 3x² while du expects 2x, never ignore the coefficient; introduce the constant fraction carefully.

一个常见错误是当导数不完全匹配时忘记调整常数因子。务必先写出 du = g'(x) dx，然后解出所需的表达式。如果被积函数中有 3x² 而 du 需要 2x，绝不可忽视系数，应谨慎引入常数分数。

Another pitfall is failing to change the limits of integration in definite integrals. Exercise 21J.1 often includes both indefinite and definite integrals; for the latter, convert the bounds to u-values or substitute back after integrating.

另一个陷阱是在定积分中忘记更换积分限。练习21J.1常同时包含不定积分和定积分；对于后者，要么将上下限转换为 u 的值，要么在积分后代回原变量再代入。

Finally, never leave the answer in terms of u. The final answer must be expressed in the original variable. In the IB exam, a response ending with u instead of x will lose the final accuracy mark.

最后，答案绝不能以 u 表示。最终结果必须用原变量表达。在IB考试中，以 u 而非 x 结尾的回答将失去最后的准确性分数。

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10. Building Speed and Fluency | 提升速度与熟练度

To ace Exercise 21J.1, practice recognising patterns without writing full substitution every time. For ∫ sin x cos x dx, you may directly think of the result (1/2)sin²x + C. However, for complex expressions, always write du and dx explicitly to avoid errors.

要攻克练习21J.1，练习在不写出完整代换过程的情况下识别模式。对于 ∫ sin x cos x dx，你可以直接想到结果 (1/2)sin²x + C。但对于复杂表达式，一定要明确写出 du 和 dx 的关系以避免错误。

Create a personal chart linking common u-choices: for f(ax+b) let u=ax+b; for √(something) let u=the radicand; for ln(something) let u=the log argument; for e^(something) let u=the exponent. This mental map will make Exercise 21J.1 feel like a sequence of familiar puzzles.

制作一个个人总结表，将常见 u 的选择连起来：对于 f(ax+b) 令 u=ax+b；对于 √(某式) 令 u=被开方式；对于 ln(某式) 令 u=对数的自变量；对于 e^(某式) 令 u=指数部分。这张脑图会让练习21J.1变得像一连串熟悉的谜题。

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11. Connecting Substitution to the Chain Rule | 将代换与链式法则联系起来

Understanding that substitution undoes the chain rule deepens mastery. If differentiating f(g(x)) gives f'(g(x))·g'(x), then integration must recover f(g(x)). Every problem in Exercise 21J.1 is secretly a derivative of some composition; thinking backward trains your intuition for integration.

理解代换是链式法则的逆运算，可以加深掌握程度。如果对 f(g(x)) 求导得到 f'(g(x))·g'(x)，那么积分必须还原出 f(g(x))。练习21J.1中的每一题本质上都是某个复合函数的导数；逆向思考能够锻炼你的积分直觉。

When stuck, mentally differentiate your candidate u and imagine what the original integral would need to become. This reverse engineering is often the quickest path to spotting the correct substitution in exam settings.

当卡住时，在心里对你预选的 u 求导，想象原积分需要变成什么样子。这种逆向工程通常是在考试环境中快速发现正确代换的最快路径。

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12. Summary and Exam Tips | 总结与考试技巧

Exercise 21J.1 is a vital stepping stone in IB Mathematics, covering polynomial, trigonometric, exponential, logarithmic, and root-type substitutions. Master these patterns, and later topics like integration by parts and trigonometric substitution will become much more manageable.

练习21J.1是IB数学中一块重要的基石，涵盖了多项式、三角、指数、对数和根式类型的代换。掌握这些模式后，后续的分部积分和三角代换等课题将变得容易得多。

In the exam, show your substitution clearly: state u, write du = … dx, rewrite the integral, integrate, and finally substitute back. Even if the final answer is wrong, clear method marks can be earned. Practice Exercise 21J.1 until you can do it with your eyes closed – the fluency will pay off across the whole calculus syllabus.

考试中，要清晰地展示代换过程：写出 u、写出 du = … dx、改写积分、积出结果、最后回代。即使最终答案有误，清晰的方法分也能拿到。反复练习练习21J.1直到闭着眼都能做出来——这种熟练度会在整个微积分大纲中给你回报。

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