Exercise 23C.2: Integration by Substitution | 练习23C.2：换元积分法

📚 Exercise 23C.2: Integration by Substitution | 练习23C.2：换元积分法

Integration by substitution is one of the most powerful and frequently used techniques for evaluating indefinite and definite integrals. In the IB Mathematics: Analysis and Approaches course, this method appears under the topic of integration, and Exercise 23C.2 provides you with the essential practice needed to master it. The core idea is to transform a complicated integral into a simpler one by changing the variable of integration. This article will walk you through the theory, step-by-step examples, common pitfalls, and a selection of worked exercises from 23C.2 to help you build confidence and accuracy.

换元积分法是求解不定积分和定积分时最常用且功能强大的技巧之一。在IB数学分析与方法的课程中，该方法属于积分章节，而练习23C.2正是帮助你掌握这一方法所必需的训练。其核心思想是通过变量代换将一个复杂的积分转化为一个简单的积分。本文将带你梳理理论、逐步拆解例题、指出常见错误，并精选23C.2中的一些习题进行讲解，帮助你建立信心、提高准确率。

1. The Substitution Rule | 换元法则

rentiable function whose range is an interval I and f is continuous on I, then ∫ f(g(x)) g'(x) dx = ∫ f(u) du. In Leibniz notation, we let u = g(x), so du/dx = g'(x), which yields du = g' the original integral into an integral in terms of u that is often much easier to handle.

换元法则指出，如果 u = g(x) 是一个可微函数，其值域为一个区间 I，且 f 在 I 上连续，那么 ∫ f(g(x)) g'(x) dx = ∫ f(u) du。用莱布尼茨记号表示，我们令 u = g(x)，则 du/dx = g'(x)，于是 du = g'(x) dx。这样，我们就能用 u 代替复合函数的内层函数，将原积分化为关于 u 的积分，后者通常更容易处理。

e integrand. When you spot something like an inside function u = h(x) and its derivative h' ikely to work. The formal statement may seem abstract, but with practice you will learn to 'see' the substitution almost instantly.

识别的关键信号是被积函数中同时出现一个函数及其导数。当你发现内层函数 u = h(x) 及其导数 h'(x) 乘上一个关于 u 的函数时，换元法很可能适用。理论的正式表述或许有些抽象，但通过练习，你很快就能近乎本能地“看”出换元路径。

2. Choosing u and du | 选取 u 与 du

Selecting an appropriate u is the most critical step. A useful guideline is to choose u as the inner function of a composite function, that is, the argument of a power, exponential, trigonometric, or logarithmic function. Once u is chosen, differentiate it to obtain du/dx, then isolate dx so that you can replace it with du. The whole original integrand in terms of x must be expressible in terms of u and du, with no x's left behind. If an x remains after substitution, you may need to solve for x in terms of u and substitute again.

选取合适的 u 是最关键的一步。一个有用的经验法则是选取复合函数的内层函数作为 u，也就是幂函数、指数函数、三角函数或对数函数内部的那个部分。选定 u 后，对它求导得到 du/dx，然后解出 dx，以便用 du 代替。原来关于 x 的被积函数整体必须能够用 u 和 du 表示，不能留下任何 x。如果代换后仍有 x 残留，你可能需要将 x 用 u 表示出来，再作一步代换。

sin(u) · (1/2) du = -½ cos(u) + C = -½ cos(x²) + C. This illustrates the complete process.

例如，在 ∫ x sin(x²) dx 中，自然的选择是 u = x²，因为它的导数 2x 出现了。于是 du = 2x dx，即 x dx = (1/2) du。积分变为 ∫ sin(u) · (1/2) du = -½ cos(u) + C = -½ cos(x²) + C。这展示了完整过程。

3. Example 1: Polynomial Composition | 例1：多项式复合

C. Practice recognizing powers and their derivatives; this structure is extremely common in Exercise 23C.2.

考虑积分 ∫ (3x² + 2x) (x³ + x²)⁵ dx。乍一看很乱，但注意到括号内表达式 x³ + x² 的导数正是 3x² + 2x。这恰好是另一个因子。令 u = x³ + x²，则 du = (3x² + 2x) dx。积分变为 ∫ u⁵ du = (1/6)u⁶ + C = (1/6)(x³ + x²)⁶ + C。练习识别幂及其导数的这种结构；这种形式在练习23C.2中极为常见。

Always adjust for constants; the presence of an extra constant does not prevent substitution.

当缺少常数因子时，会出现变体。对于 ∫ x² (x³ + 1)⁴ dx，我们设 u = x³ + 1，du = 3x² dx，则 x² dx = (1/3) du。然后 ∫ (1/3) u⁴ du = (1/15) u⁵ + C。始终要为常数进行调整；多出一个常数并不妨碍换元。

4. Example 2: Trigonometric Substitution | 例2：三角函数代换

t the first choice is cleaner because the odd power of cosine lets us save one cosine for du. The general strategy is to save one factor of the derivative of your chosen u.

三角函数的积分经常需要换元。计算 ∫ cos³(x) sin(x) dx。令 u = cos(x)，则 du = -sin(x) dx，于是 sin(x) dx = -du。积分变为 ∫ u³ (-du) = -∫ u³ du = -(1/4)u⁴ + C = -¼ cos⁴(x) + C。你也可以设 u = sin(x)，但第一种选择更简洁，因为 cos 的奇次幂允许我们保留一个余弦作为 du。一般策略是为你选择的 u 保留一个它的导数因子。

is ∫ tan(x) dx = ∫ sin(x)/cos(x) dx. Let u = cos(x), du = -sin(x) dx, giving -∫ du/u = -ln|u| + C = -ln|cos(x)| + C = ln|sec(x)| + C. This substitution appears repeatedly, so memorise the approach.

另一个经典例子是 ∫ tan(x) dx = ∫ sin(x)/cos(x) dx。令 u = cos(x)，du = -sin(x) dx，得到 -∫ du/u = -ln|u| + C = -ln|cos(x)| + C = ln|sec(x)| + C。这种代换反复出现，建议记住这一方法。

5. Example 3: Exponential Functions | 例3：指数函数

ightforward, but when the exponent is a function, substitution is required. Evaluate ∫ x e^x² dx. Let u = x², du = 2x dx, so x dx = (1/2) du. Then ∫ (1/2) e^u du = ½ e^u + C = ½ e^x² + C. This pattern is frequent in earlier exercises.

指数为线性的指数函数积分很简单，但当指数本身是一个函数时，就需要换元。计算 ∫ x e^x² dx。令 u = x²，du = 2x dx，则 x dx = (1/2) du。于是 ∫ (1/2) e^u du = ½ e^u + C = ½ e^x² + C。这种模式在之前的练习中频繁出现。

Sometimes the substitution is less obvious, as in ∫ e^√x / √x dx. Let u = √x, so du = 1/(2√x) dx, meaning (1/√x) dx = 2 du. The integral becomes 2 ∫ e^u du = 2e^u + C = 2e^√x + C. Recognising the derivative of the exponent inside the integrand is the skill to hone.

有时代换不那么明显，例如 ∫ e^√x / √x dx。令 u = √x，则 du = 1/(2√x) dx，即 (1/√x) dx = 2 du。积分变为 2 ∫ e^u du = 2e^u + C = 2e^√x + C。识别出被积函数中指数部分的导数，这是需要磨练的技能。

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6. Example 4: Rational Functions | 例4：有理函数

+ 3x + 5) dx, set u = x² + 3x + 5, then du = (2x + 3) dx. The integral becomes ∫ du/u = ln|u| + C = ln|x² + 3x + 5| + C. This is a direct application of the ∫ f'(x)/f(x) dx = ln|f(x)| + C rule, which itself is a special case of substitution.

分子是分母导数的（或倍数）的有理函数，非常适合对数换元。对于 ∫ (2x + 3)/(x² + 3x + 5) dx，令 u = x² + 3x + 5，则 du = (2x + 3) dx。积分变为 ∫ du/u = ln|u| + C = ln|x² + 3x + 5| + C。这是 ∫ f'(x)/f(x) dx = ln|f(x)| + C 的直接应用，而该法则本身就是换元法的特例。

u = x+1, du = dx. The integral becomes ∫ du/(u² + 2²) = (1/2) arctan(u/2) + C. Such problems often appear in Exercise 23C.2 to test your versatility.

部分分式是另一种标准方法，但当分母是实数范围内不可分解的二次式时，可能需要通过配方并使用反正切形式的代换。例如，∫ 1/(x² + 2x + 5) dx。将分母写成 (x+1)² + 4，然后令 u = x+1，du = dx。积分变为 ∫ du/(u² + 2²) = (1/2) arctan(u/2) + C。这类问题经常在练习23C.2中出现，以考察你综合运用的能力。

7. Definite Integrals and Substitution | 定积分与换元

, then u = g(x) goes from g(a) to g(b). After substitution, you never return to x; you simply integrate with respect to u and apply the new limits.

在定积分中使用换元法时，你有两种选择。选项一：求出关于 u 的原函数，然后换回 x，并在原积分上下限处求值。选项二（通常更优）：在换元的同时改变积分上下限。若 x 从 a 到 b，那么 u = g(x) 就从 g(a) 到 g(b)。代换之后，你无需再回到 x；直接对 u 积分并代入新的上下限即可。

For example, evaluate ∫₀^√π dx. Let u = x², du = 2x dx. When x = 0, u = 0; when x = √π, u = π. The integral becomes ∫₀^π cos(u) du = [sin(u)]₀^π = sin(π) - sin(0) = 0. This method saves time and reduces algebraic mistakes.

例如，计算 ∫₀^√π 2x cos(x²) dx。令 u = x²，du = 2x dx。当 x = 0 时，u = 0；当 x = √π 时，u = π。积分变为 ∫₀^π cos(u) du = [sin(u)]₀^π = sin(π) - sin(0) = 0。这一方法节省时间并减少了代数错误。

Be careful with definite integrals: always check whether the substitution is one-to-one over the interval to avoid issues with the change‑of‑limits formula. For the standard functions in Exercise 23C.2, this is typically satisfied.

处理定积分时要小心：务必检验在积分区间上代换是否是一一映射，以避免在变换积分限时出现问题。对于练习23C.2中的标准函数，此条件通常都满足。

8. Common Pitfalls and How to Avoid Them | 常见错误及避免方法

to replace dx with du correctly, especially missing a constant factor. Always write du = g'(x) dx, then solve for dx explicitly: dx = du / g' s in terms of u. If you still see an x, you must eliminate it using the relation u = g(x).

最常见的错误是忘记正确地将 dx 替换为 du，特别是漏掉常数因子。一定要写出 du = g'(x) dx，然后明确解出 dx：dx = du / g'(x)。代换之后，检查整个被积函数是否全都用 u 表示了。如果还能看到 x，就必须利用关系式 u = g(x) 将其消去。

Another pitfall is differentiating u incorrectly, especially with trigonometric or exponential functions. Double-check your derivative before proceeding. Also, remember the absolute value inside logarithms for integrals yielding ln, unless the domain guarantees positivity. Finally, when doing back‑substitution, ensure you replace u with the original expression and simplify.

另一个易错点是求 u 的导数时出错，尤其是涉及三角或指数函数时。在进行下一步之前，再次检查求导结果。另外，对于得到 ln 的积分，要记住对数里加绝对值，除非定义域保证了正值。最后，在回代时，确保用原表达式替换 u 并化简。

For definite integrals, students often forget to change the limits. If you convert back to x and use the old limits, the result is correct but the work is longer. If you keep everything in u, you must apply the new limits. Either way, be consistent.

对于定积分，学生常常忘记变换积分限。如果换回 x 并使用原来的上下限，结果是正确的，但过程更繁琐。如果保持 u 的形式，就必须使用新的上下限。无论采用哪种方式，都要保持一致。

9. Worked Examples from Exercise 23C.2 | 练习23C.2精选例题详解

Let us now apply these ideas to a few problems typical of Exercise 23C.2. Problem 1: ∫ x √2x² + 1 dx. Choose u = 2x² + 1, du = 4x dx, so x dx = (1/4) du. Integral: ∫ √u · (1/4) du = (1/4) · (2/3) u^3/2 + C = (1/6)(2x² + 1)^3/2 + C. This illustrates handling square roots.

现在，我们将这些思路应用于练习23C.2中的几个典型问题。问题1：∫ x √2x² + 1 dx。选择 u = 2x² + 1，du = 4x dx，所以 x dx = (1/4) du。积分：∫ √u · (1/4) du = (1/4) · (2/3) u^3/2 + C = (1/6)(2x² + 1)^3/2 + C。这展示了平方根的处理方式。

= -cos(ln x) + C. Here the substitution is straightforward once you notice the 1/x factor.

问题2：∫ sin(ln x) / x dx。令 u = ln x，du = (1/x) dx。那么积分变为 ∫ sin(u) du = -cos(u) + C = -cos(ln x) + C。一旦你注意到 1/x 因子，这里的代换就很直接了。

Problem 3 (definite): ∫₁^e² dx. Set u = ln x, du = (1/x) dx. When x = 1, u = 0; when x = e², u = 2. Integral becomes ∫₀² u² du = [u³/3]₀² = 8/3. This is a classic IB examination style.

问题3（定积分）：∫₁^e² (ln x)² / x dx。设 u = ln x，du = (1/x) dx。当 x = 1 时，u = 0；当 x = e² 时，u = 2。积分变为 ∫₀² u² du = [u³/3]₀² = 8/3。这是典型的IB考试风格。

Problem 4: ∫ e^{sin x} cos x dx. Let u = sin x, du = cos x dx. The integral is ∫ e^u du = e^u + C = e^{sin x} + C. Identifying cos x dx as the derivative of sin x makes this trivial.

问题4：∫ e^{sin x} cos x dx。令 u = sin x，du = cos x dx。积分为 ∫ e^u du = e^u + C = e^{sin x} + C。识别出 cos x dx 是 sin x 的导数，使问题变得简单。

Continue working through your set; each problem reinforces pattern recognition. If you get stuck, ask yourself: 'What is the inner function? What is its derivative? Do I see that derivative in the integrand?'

继续完成你的练习集；每道题目都在强化模式识别能力。如果遇到困难，问自己：“内层函数是什么？它的导数是什么？我在被积函数中看到该导数了吗？”

10. Summary and Final Tips | 总结与最后建议

Integration by substitution is a cornerstone of integral calculus. Mastery comes from recognising the relationship between a function and its derivative inside the integrand. Always explicitly write u, compute du, and adjust for any missing constants. Practice both indefinite and definite integrals, and for the latter, get comfortable with changing limits. In the IB exam, substitution is often the first step in a multi‑step problem, so fluency here sets you up for success in more advanced topics like integration by parts, differential equations, and area/volume calculations.

换元积分法是积分学的基石。能否掌握精髓，取决于能否识别出被积函数中函数与其导数之间的关系。始终要明确写出 u，计算 du，并对任何缺失的常数进行调整。练习不定积分和定积分，对于后者，要熟练掌握变换积分限的方法。在IB考试中，换元往往是多步问题的第一步，因此在此处达到熟练掌握，将为你学习分部积分、微分方程以及面积/体积计算等更高级的专题奠定成功基础。

When tackling Exercise 23C.2, work systematically, check your answers by differentiation, and don't rush. The time you invest now in careful practice will pay off enormously in your internal assessment and final examinations.

在攻克练习23C.2时，要系统地进行，用求导检验你的答案，不要急躁。你现在在精细练习上投入的时间，将会在内部评估和期末考试中获得巨大回报。

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