📚 Exercise 17A.2: Integration by Substitution | 练习 17A.2：换元积分法
Integration by substitution, often called the reverse chain rule, is one of the most powerful tools in IB Mathematics. Exercise 17A.2 is designed to build fluency in recognising the inner function, differentiating it, and rewriting the integral entirely in terms of a new variable. Mastering this technique is essential for success in both Analysis and Approaches and Applications and Interpretation courses.
换元积分法，常被称为反向链式法则，是 IB 数学中最强大的工具之一。练习 17A.2 旨在帮助学生熟练识别内层函数、对其求导，并将积分完全改写为新变量的表达式。掌握这一技巧对于在分析与方法和应用与解释两门课程中取得好成绩至关重要。
1. Understanding Integration by Substitution | 理解换元积分法
d by the derivative of g(x), we can simplify the integral by letting u = g(x). Then du = g'(x)dx, and the integral becomes ∫ f(u) du, which is often straightforward to evaluate.
换元积分法通过反向运用微分中的链式法则实现。当被积函数包含复合函数 f(g(x)) 并与 g(x) 的导数相乘时，我们可以设 u = g(x) 来简化积分。于是 du = g'(x)dx，积分就转化为 ∫ f(u) du，通常很容易计算。
2. The Reverse Chain Rule Formula | 反向链式法则公式
The formal pattern that governs Exercise 17A.2 is ∫ f'(g(x))·g'(x) dx = f(g(x)) + C. In practice, we set u = g(x), compute du = g' to u, we substitute back to express the final answer in terms of the original variable x.
练习 17A.2 的形式化模式为 ∫ f'(g(x))·g'(x) dx = f(g(x)) + C。实际操作中，我们设 u = g(x)，计算 du = g'(x)dx，并将积分完全用 u 表示。对 u 积分后，再回代以用原始变量 x 表示最终答案。
∫ f(g(x))·g'(x) dx = ∫ f(u) du = F(u) + C = F(g(x)) + C
∫ f(g(x))·g'(x) dx = ∫ f(u) du = F(u) + C = F(g(x)) + C
3. Choosing the Right Substitution u | 选择合适的代换 u
function. In Exercise 17A.2 you will often see u = ax + b, u = sin x, u = x², or u = 1 + eˣ.
选择代换是最关键的一步。寻找一个其导数也出现在被积函数中的函数。典型的选择包括幂次中的线性表达式、三角函数的参数、指数中的表达式，或分母中的函数。在练习 17A.2 中，你经常会看到 u = ax + b、u = sin x、u = x² 或 u = 1 + eˣ。
4. Worked Example 1: ∫ (2x+1)³ dx | 例题 1：∫ (2x+1)³ dx
du = (1/2) · (u⁴/4) + C = u⁴/8 + C. Finally, replace u with 2x+1 to obtain (2x+1)⁴/8 + C.
设 u = 2x+1。则 du/dx = 2，于是 dx = du/2。全部代入：∫ (2x+1)³ dx = ∫ u³ · (du/2) = (1/2) ∫ u³ du = (1/2) · (u⁴/4) + C = u⁴/8 + C。最后将 u 换回 2x+1，得到 (2x+1)⁴/8 + C。
5. Worked Example 2: ∫ x·sin(x²) dx | 例题 2：∫ x·sin(x²) dx
(du/2) = (1/2) ∫ sin u du = -(1/2) cos u + C. Back-substituting gives -(1/2) cos(x²) + C.
注意到 x² 的导数是 2x，与前面的 x 因子紧密相关。设 u = x²，则 du/dx = 2x，即 x dx = du/2。积分变为 ∫ sin(u) · (du/2) = (1/2) ∫ sin u du = -(1/2) cos u + C。回代得到 -(1/2) cos(x²) + C。
6. Worked Example 3: ∫ eˣ/(1+eˣ) dx | 例题 3：∫ eˣ/(1+eˣ) dx
Here the numerator eˣ is exactly the derivative of the denominator 1+eˣ. Let u = 1+eˣ, then du = eˣ dx. The integral simplifies to ∫ 1/u du = ln|u| + C. Since 1+eˣ is always positive, the absolute value can be dropped, yielding ln(1+eˣ) + C.
此处分子的 eˣ 恰好是分母 1+eˣ 的导数。令 u = 1+eˣ，则 du = eˣ dx。积分简化为 ∫ 1/u du = ln|u| + C。由于 1+eˣ 恒为正，绝对值可省略，得到 ln(1+eˣ) + C。
7. Substitution with Definite Integrals | 定积分中的换元法
e new variable u or substitute back to x before applying the original limits. For ∫₀¹ 2x ex² du = 2x dx. Transform the limits: when x=0, u=0; when x=1, u=1. The integral becomes ∫₀¹ eu du = [eu]₀¹ = e - 1. Never mix variables with inappropriate limits.
处理定积分时，必须将积分上下限也换为新变量 u，或者在回代后再应用原上下限。对 ∫₀¹ 2x ex² dx，令 u = x²，则 du = 2x dx。变换上下限：x=0 时 u=0；x=1 时 u=1。积分变为 ∫₀¹ eu du = [eu]₀¹ = e - 1。切勿混用变量与不匹配的上下限。
8. Common Mistakes and Pitfalls | 常见错误与陷阱
- ing to replace dx with du correctly, e.g. writing dx = du but missing the required factor.
- 忘记将 dx 正确地换成 du，例如写 dx = du 却遗漏了所需的因子。
- Choosing a substitution that does not simplify the integral, leaving the integrand still a mixture of x and u.
- 选择的代换没有简化积分，导致被积函数仍混合着 x 和 u。
- Neglecting to back-substitute u in indefinite integrals, leaving the answer in terms of u instead of x.
- 在不定积分中忘记回代 u，导致答案用 u 而不是 x 表示。
- In definite integrals, applying x-limits to a u-integral without converting the limits.
- 在定积分中，直接对 u 积分套用 x 的上下限而不进行转换。
9. Tips for IB Exam Success | IB 考试成功技巧
- explicitly before substituting. This helps examiners follow your work and reduces errors.
- 在代换前务必清晰地写出 u = ... 和 du = ...。这有助于考官理解你的解题过程，并减少错误。
- Check your answer by differentiating. If you obtain the original integrand, your integration is correct.
- 通过求导来检验答案。如果得到原被积函数，则积分正确。
- Practice recognising patterns: linear inside a power, trig(ax+b), exponential with a multiple, and rational functions where the numerator is close to the denominator's derivative.
- 练习识别模式：幂内部的线性式、sin/cos(ax+b)、带倍数的指数，以及分子接近分母导数的有理函数。
- For definite integrals, change limits immediately and mark them clearly on your work.
- 对于定积分，立即转换上下限并清晰标注。
10. Practice Problems in the Style of Exercise 17A.2 | 练习 17A.2 风格的习题
Apply the substitution method to the following problems. Solve each step by step, and then check your results by differentiation.
对以下题目应用换元法。逐步求解，然后通过求导检验结果。
1. ∫ (3x-2)⁴ dx
1. ∫ (3x-2)⁴ dx
2. ∫ cos(5x) dx
2. ∫ cos(5x) dx
3. ∫ x² √x³+1 dx
3. ∫ x² √x³+1 dx
4. ∫ (ln x)/x dx for x > 0
4. ∫ (ln x)/x dx，x > 0
5. ∫₀½ 1/(1-x) dx
5. ∫₀½ 1/(1-x) dx
11. Extension: Substitution and Inverse Trigonometric Functions | 拓展：换元法与反三角函数
pproached by recognising it as the derivative of arctan x. A more advanced pattern is ∫ 1/√a²-x² itution like x = a sin θ, which is a further development of the skills practised in 17A.2.
换元法也出现在导出反三角函数形式的积分中。例如，∫ 1/(1+x²) dx 可直接识别为 arctan x 的导数。更复杂的模式如 ∫ 1/√a²-x² dx，常使用三角代换 x = a sin θ，这是在练习 17A.2 基础上的进一步发展。
12. Summary and Key Takeaways | 总结与核心要点
- Identify the inner function g(x) whose derivative g'(x) appears as a factor.
- 识别内层函数 g(x)，其导数 g'(x) 作为因子出现。
- Set u = g(x), compute du = g'(x) dx, and rewrite the integral solely in terms of u.
- 设 u = g(x)，计算 du = g'(x) dx，并仅用 u 改写积分。
- Integrate with respect to u, then substitute back or adjust limits for definite integrals.
- 对 u 积分，然后回代或调整定积分上下限。
- Always verify by differentiation and avoid common pitfalls related to dx substitution and limit conversion.
- 务必通过求导验证，并避免与 dx 代换和上下限转换相关的常见错误。
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图