Exercise 21J.3: Integration by Substitution | 练习 21J.3：换元积分法

📚 Exercise 21J.3: Integration by Substitution | 练习 21J.3：换元积分法

This article unpacks a classic IB Mathematics exercise on integration by substitution, a technique that reverses the chain rule. We work through Exercise 21J.3 step by step, exploring the reasoning, routine, and common pitfalls. By the end, you will be confident in handling integrals involving composite functions and matching differentials.

本文剖析一个经典的 IB 数学换元积分练习题，该技巧本质上是链式法则的逆运算。我们将逐步推演练习 21J.3，解析其中的推理、计算流程与常见误区。读完本文，你将对处理含复合函数并配齐微分的积分充满信心。

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1. Understanding the Substitution Method | 理解代换积分法

Integration by substitution is the reverse of the chain rule for differentiation. If an integrand can be written as f(g(x))·g'(x), then letting u = g(x) transforms the integral into ∫ f(u) du, which is often easier to evaluate. This method is especially powerful when the derivative of an inner function appears as a factor.

代换积分法是复合函数求导（链式法则）的逆过程。如果被积函数可以写成 f(g(x))·g'(x) 的形式，令 u = g(x) 就能把积分化为 ∫ f(u) du，通常更容易计算。当内层函数的导数作为因子出现时，这个方法特别有效。

In IB Mathematics, you will encounter substitution in contexts ranging from polynomial expressions to trigonometric, exponential, and logarithmic integrals. Recognising the pattern “function and its derivative” is key.

在 IB 数学中，你会遇到代换法处理多项式、三角函数、指数和对数等积分。识别“函数与其导数共存”的模式是关键。

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2. Problem Statement: Exercise 21J.3 | 题目描述：练习 21J.3

The exercise presents the following indefinite integral: ∫ x √(x² + 1) dx. At first glance, it looks messy because of the square root of a quadratic, but the extra x factor is a strong hint for substitution.

练习给出如下不定积分：∫ x √(x² + 1) dx。初看因包含二次式的平方根而显得有些复杂，但多出的 x 因子强烈暗示可以使用代换。

Our task is to evaluate this integral accurately and express the result in terms of the original variable x, including the constant of integration.

我们的任务是准确计算这个积分，并用原变量 x 表示结果，同时包含积分常数。

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3. Step 1: Identify the Inner Function | 第一步：识别内层函数

Look inside the composition: the expression x² + 1 sits under the square root. It is natural to set u = x² + 1. This inner function is simple, its derivative is 2x, and the integrand already contains x dx, which is almost du/2.

观察复合结构：表达式 x² + 1 位于平方根内。自然设定 u = x² + 1。这个内层函数简单，其导数为 2x，而被积函数中恰好有 x dx，几乎就是 du/2。

This choice is strategic because it collapses √(x² + 1) into √u and allows us to replace the remaining x dx. Always ask: does the derivative of the chosen u appear in the integrand? Here, yes – to within a constant factor.

这种选择策略性强，因为它将 √(x² + 1) 压缩为 √u，并容许替换剩余的 x dx。始终要问：所选的 u 的导数是否出现在被积函数中？这里正是如此——只差一个常数因子。

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4. Step 2: Compute du/dx and Rearrange | 第二步：计算 du/dx 并整理

Differentiate u with respect to x: du/dx = 2x. In differential form, this is du = 2x dx. Rearranging to isolate the x dx term that appears in our integral gives x dx = du / 2 or equivalently ½ du.

对 u 关于 x 求导：du/dx = 2x。写成微分形式即 du = 2x dx。重新整理以分离出我们积分中出现的 x dx 项，得到 x dx = du / 2，即 ½ du。

This rearrangement is crucial because it replaces everything in the original integrand with expressions in u. Missing this step often leads to errors where dx is incorrectly substituted.

这一整理至关重要，因为它将原被积函数中的所有部分都替换为关于 u 的表达式。遗漏这一步常导致 dx 替换错误。

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5. Step 3: Substitute into the Integral | 第三步：代入积分

Replace √(x² + 1) with √u, and x dx with ½ du. The integral becomes ∫ √u · (½ du) = ½ ∫ u^(1/2) du. Note that the constant factor ½ can be taken outside the integral sign.

用 √u 替换 √(x² + 1)，用 ½ du 替换 x dx。积分变为 ∫ √u · (½ du) = ½ ∫ u^(1/2) du。注意常数因子 ½ 可以提到积分号外面。

At this stage, we have transformed a somewhat complicated integrand into a simple power of u. The role of substitution is complete: the hard work is done.

至此，我们已经把稍微复杂的被积函数转化成了简单的 u 的幂函数。代换的任务已完成：困难的部分解决了。

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6. Step 4: Integrate with Respect to u | 第四步：对 u 积分

Use the power rule for integration: ∫ u^n du = u^(n+1)/(n+1) + C, provided n ≠ −1. Here n = 1/2, so n+1 = 3/2. Thus ∫ u^(1/2) du = u^(3/2) / (3/2) + C = (2/3) u^(3/2) + C.

运用幂函数的积分法则：∫ u^n du = u^(n+1)/(n+1) + C，前提是 n ≠ −1。此处 n = 1/2，因而 n+1 = 3/2。于是 ∫ u^(1/2) du = u^(3/2) / (3/2) + C = (2/3) u^(3/2) + C。

Multiply by the factor ½ from outside: ½ × (2/3) u^(3/2) = (1/3) u^(3/2) + C. The constant C absorbs any numerical constant, so we can simply write + C once.

乘以外部的因子 ½：½ × (2/3) u^(3/2) = (1/3) u^(3/2) + C。常数 C 吸收所有数值常量，因此最后只写一次 + C 即可。

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7. Step 5: Back-Substitute to x | 第五步：代回原变量 x

Replace u with the original expression x² + 1. The final antiderivative in terms of x is (1/3) (x² + 1)^(3/2) + C. Always present the answer in the variable you started with unless instructions specify otherwise.

将 u 换回原表达式 x² + 1。关于 x 的最终原函数是 (1/3) (x² + 1)^(3/2) + C。除非题目另有要求，永远用起始变量表示答案。

This result represents a family of functions whose derivative is x √(x² + 1). It is customary to leave the power (3/2) as an exponent rather than converting to a radical, though both forms are acceptable.

这个结果表示导数为 x √(x² + 1) 的一族函数。习惯上保留 (3/2) 次幂的形式而不转换回根号，虽然两种形式都可以。

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8. Step 6: Verify by Differentiation (Optional but Recommended) | 第六步：通过求导验证（可选但推荐）

Differentiate (1/3)(x² + 1)^(3/2) using the chain rule. The derivative is (1/3) × (3/2)(x² + 1)^(1/2) × (2x) = (1/2)(2x)√(x² + 1) = x √(x² + 1). The original integrand is recovered, confirming correctness.

对 (1/3)(x² + 1)^(3/2) 用链式法则求导。导数为 (1/3) × (3/2)(x² + 1)^(1/2) × (2x) = (1/2)(2x)√(x² + 1) = x √(x² + 1)。恢复了原始被积函数，确认无误。

This quick check is a powerful habit for exams: it can catch algebraic slips and reinforce the connection between integration and differentiation.

这一快速检验是考试中的一个有力习惯：它可以捕获代数失误，并强化积分与求导之间的联系。

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9. Common Errors to Avoid | 常见错误与规避

Forgetting to adjust dx: The most frequent mistake is writing du = dx or omitting the relationship between du and x dx. Always express the original differential completely in terms of u and du.

忘记调整 dx：最常见的错误是写成 du = dx 或忽略 du 与 x dx 之间的关系。务必将原微分完全用 u 和 du 表示。

Incorrect reverse substitution: After integrating, students sometimes leave the answer in u. The IB mark scheme will only award full marks if the final answer is given in x, unless the question states otherwise.

代回错误：积分后，学生有时会留下关于 u 的答案。IB 评分方案只在最终答案用 x 表示时才给满分，除非题设另有说明。

Ignoring the constant factor: When the derivative is missing a constant multiple, some learners fail to compensate. Here, the derivative contributed a factor of 2, which we correctly handled by introducing ½.

忽略常数因子：当导数缺少一个常数倍时，部分学习者未能补偿。此题中导数带来了因子 2，我们通过引入 ½ 正确处理了。

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10. Extension: Definite Integral Version | 拓展：定积分情形

Suppose the exercise instead asked for the definite integral from x = 0 to x = 1 of the same integrand. You have two approaches: transform the limits along with the variable, or back-substitute before evaluating. Both should yield the same result: (1/3)(2^(3/2) − 1).

假设练习改为求同一被积函数从 x = 0 到 x = 1 的定积分。有两种方式：变换积分限（随变量代换），或先代回原变量再代值。两种方法都应给出相同结果：(1/3)(2^(3/2) − 1)。

If you change the limits, when u = x² + 1: x = 0 gives u = 1; x = 1 gives u = 2. The integral becomes ½ ∫_{1}^{2} u^(1/2) du, which evaluates neatly. Using original limits avoids back-substitution but requires careful limit conversion.

若改变积分限，由 u = x² + 1：x = 0 时 u = 1；x = 1 时 u = 2。积分变为 ½ ∫_{1}^{2} u^(1/2) du，计算整洁。使用原始积分限可以避免回代，但需谨慎转换上下限。

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11. Recognising Substitution Patterns | 识别代换模式

Exercise 21J.3 belongs to a broader family: integrals of the form ∫ f'(x)·g(f(x)) dx. In our case, f(x) = x² + 1 and g(u) = √u. With practice, you will spot these structures instantly.

练习 21J.3 属于更广泛的族类：形如 ∫ f'(x)·g(f(x)) dx 的积分。此处 f(x) = x² + 1，g(u) = √u。通过练习，你将能瞬间识别这些结构。

Pattern (模式)	Substitution (代换)	Example (示例)
∫ (ax + b)ⁿ dx, with derivative mismatch	u = ax + b	∫ x (2x+3)⁵ dx
∫ f'(x)/f(x) dx	u = f(x)	∫ (2x)/(x²+5) dx
∫ f'(x)·√f(x) dx	u = f(x)	∫ x √(x²+1) dx
∫ f'(x)·e^(f(x)) dx	u = f(x)	∫ 2x e^(x²) dx
∫ sinⁿx cos x dx	u = sin x	∫ sin²x cos x dx

Memorising these common forms helps you decide on a substitution quickly, saving valuable time in an IB examination.

记忆这些常见形式有助于快速选定代换变量，在 IB 考试中节省宝贵时间。

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12. Practice Exercises and Final Tips | 配套练习与最终建议

To consolidate your skills, attempt these similar indefinite integrals:

为巩固技能，请尝试以下类似的不定积分：

• ∫ x² √(x³ + 2) dx
• 中文：∫ x² √(x³ + 2) dx

• ∫ (ln x) / x dx
• 中文：∫ (ln x) / x dx

• ∫ e^(sin x) cos x dx
• 中文：∫ e^(sin x) cos x dx

For each, identify the inner function, compute du, perform substitution, integrate, and back-substitute. Compare your steps with the structured approach demonstrated above.

对每一题，识别内层函数，计算 du，进行代换，积分并代回。将你的步骤与上面展示的结构化方法进行比较。

Finally, remember that integration by substitution is a tool, not a trick. Understand why it works—the reverse chain rule—and you will find it intuitive rather than mechanical.

最后，请记住换元积分是一种工具，而非技巧。理解其原理——链式法则的逆向运用——你就会觉得它直观而不是机械操作。

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