📚 Integration by Substitution | 换元积分法
Integration by substitution is one of the core techniques in A-Level Edexcel Pure Mathematics, enabling you to simplify complicated integrals by replacing a function of x with a single variable u. This method reverses the chain rule for differentiation and is widely tested in both the pure and applied papers. Mastering substitution not only builds algebraic fluency but also lays the groundwork for further calculus topics in higher education.
换元积分法是 A-Level Edexcel 纯数学的核心技巧之一，通过将含 x 的函数替换为单一变量 u，可以大幅简化复杂的积分表达式。该方法本质上是链式法则的逆用，在纯数及应用试卷中频繁出现。熟练掌握换元法不仅能提升代数运算流畅度，也为高等数学中的进阶微积分学习奠定基础。
1. The Basic Principle | 基本原理
ginal variable x and differential dx are then completely replaced by expressions in u and du. After integration with respect to u, the final answer is switched back to x. The reverse chain rule is the intuitive backbone: if d/dx [F(g(x))] = F'(g(x)) g'(x), then integrating F'(g(x)) g' (x)). Substitution formalises this logic when the inner function g(x) and its derivative g'(x) are not immediately obvious.
换元积分法本质上通过选择合适的代换 u = g(x)，使被积函数变为关于 u 的函数并乘以 du/dx。原变量 x 和微分 dx 随即被完全替换为 u 及 du 的表达式。关于 u 积分完成后，再把结果换回 x。逆向链式法则提供了直观支撑：如果 d/dx [F(g(x))] = F'(g(x)) g'(x)，则直接对 F'(g(x)) g'(x) 关于 x 积分便得到 F(g(x))。当内层函数 g(x) 及其导数 g'(x) 不直接显式时，换元法正式将这一逻辑系统化。
2. Step-by-Step Substitution Procedure | 换元步骤拆解
When facing an integral, first scan for a composite structure — /dx, then rearrange to express dx as du / (du/dx). Substitute all x terms, including any leftover x factors, in terms of u using the original substitution. Simplify the resulting integral entirely in terms of u, perform the integration, and finally substitute back u = g(x) to obtain the answer in terms of x. Always remember to adjust limits immediately if this is a definite integral.
面对积分时，首先寻找复合结构——一个函数嵌套在另一个函数中，常见于幂次、三角函数、指数或分母内部。将内层函数记为 u。接着计算 du/dx，并重新整理成 dx = du / (du/dx) 的形式。利用原始代换关系，把所有 x 项（包括剩余的 x 因子）都换成 u 的表达式。将积分完全简化成仅含 u 的形式，完成积分，最后把 u = g(x) 回代得到关于 x 的答案。若是定积分，切记第一时间把积分上下限也一并替换。
3. Choosing the Right u | 选取合适的 u
s elsewhere in the integrand, or at least a constant multiple of it. For example, in ∫ x ex² unctions like (3x+2)⁵, u = 3x+2 simplifies the power while dx = du/3 absorbs the constant factor effortlessly.
换元能否成功很大程度上取决于 u 的选择是否正确。一个通用原则：令 u 为其导数也在被积式中出现（或为常数倍数）的那部分。例如，在 ∫ x ex² dx 中，选择 u = x² 很合适，因为 du/dx = 2x 正好提供了匹配的 x dx。在 ∫ sin³x cos x dx 中，u = sin x 是理想选择，因为 du = cos x dx。当被积函数含有线性内层函数，如 (3x+2)⁵ 时，令 u = 3x+2 可以简化幂次，同时 dx = du/3 轻松吸收常数因子。
4. Handling Definite Integrals with Limits | 处理定积分的上下限
nitely, revert to x, and then apply the original limits. The first method is usually cleaner. Replace each x‑limit by substituting into u = g(x) to obtain the new u‑limits. Then the entire integral, including limits, is in u — never return to x. For instance, if the x‑ y become 1 and 2. Be meticulous with signs when substitution reverses the order of limits.
对于定积分，有两种处理方式：要么换元同时将上下限替换，直接在 u 变量下求值；要么先不定积分，回代 x 后再代入原限制。通常第一种方法更清晰。将每个 x 上下限代入 u = g(x) 得到新的 u 上下限。这样整个积分连同积分限都处于 u 变量下——无需再回到 x。例如，若 x 上下限为 0 和 1，且 u = x²，则新限仍为 0 和 1；但若 u = x+1，则新限变为 1 和 2。当替换导致积分上下限顺序颠倒时，务必谨慎处理符号。
5. Substitution for Rational Functions | 有理函数的换元
Integrals of the form ∫ f'(x)/f(x) dx directly yield ln|f(x)| + C, which is a special one‑ n like u = linear denominator often simplifies. Consider ∫ (2x)/(x²+1) dx: u = x²+1 gives du = 2x dx, converting the integral to ∫ 1/u du = ln|u| + C. For partial‑fraction‑ ns, substitution may be used before or alongside decomposition. Always check if a simple u‑sub can reduce the degree of the denominator before applying heavy algebraic methods.
形如 ∫ f'(x)/f(x) dx 的积分直接得到 ln|f(x)| + C，这实际上是一步到位的换元特例。更一般地，对于分子并非恰为分母导数的有理式，可尝试令 u = 一次分母以简化。考虑 ∫ (2x)/(x²+1) dx：令 u = x²+1 则 du = 2x dx，积分转化为 ∫ 1/u du = ln|u| + C。对于需要部分分式分解的表达式，换元可提前使用或配合分解进行。在动用繁重代数方法之前，总是先检查一个简单的 u 换元能否降低分母次数。
6. Trigonmetric Substitutions | 三角换元
Special substitutions using sine, tangent, or secant are required when the integrand contains terms like √a²−x², √a²+x², or √x²−a². Edexcel A‑Level mainly expects you to recognise given substitutions, e.g., “use x = sin θ” to simplify √1−x² θ), the root simplifies, and the resulting trig integral is evaluated. Always draw a right‑angle triangle to return to x: if x = a sin θ, then cos θ = √a²−x²/a. This visual aid prevents algebraic errors when back‑substituting.
当被积函数包含 √a²−x²、√a²+x² 或 √x²−a² 等形式时，需要使用正弦、正切或正割等特殊换元。在 Edexcel A‑Level 考纲中，主要要求学生识别给定的换元，例如“利用 x = sin θ 将 √1−x² 化为 cos θ”。三角换元后，dx 变为 cos θ dθ（当 x = sin θ 时），根式化简，然后计算得到的三角积分。回代 x 时，务必绘制直角三角形：若 x = a sin θ，则 cos θ = √a²−x²/a。这一视觉辅助可有效避免回代时的代数错误。
7. Exponential and Logarithmic Substitution | 指数与对数换元
When the integrand contains eax+b, letting u = ax+b (the exponent) converts dx to du/a and often removes the exponential's linear argument. For integrals mixing exponentials with polynomials, such as ∫ x ex² dx, a polynomial substitution like u = x² works perfectly. In some cases, letting u = ex itself can turn a rational function in ex into a rational function in u, because du = ex dx and dx = du/u. This technique is also useful for hyperbolic‑ in further maths, but in Core Pure it helps with certain exponential rational expressions.
当被积函数含有 eax+b 时，令 u = ax+b（指数部分）可将 dx 转化为 du/a，并常常消去指数中的线性参数。对于指数与多项式混合的积分，如 ∫ x ex² dx，采用多项式换元 u = x² 效果很好。在某些情况下，直接令 u = ex 本身能把关于 ex 的有理函数变成关于 u 的有理函数，因为 du = ex dx 且 dx = du/u。这一技巧在进阶数学的双曲型积分中也十分有用，但在纯数核心里可帮助处理特定的指数有理式。
8. Integration by Substitution vs. Standard Patterns | 换元积分与标准模式的对比
+ C follows directly from differentiation of sin(ax+b). Using substitution here (u = ax+b) reinforces the pattern but is often skipped in timed exams. Similarly, ∫ f'(x) [f(x)]ⁿ dx = [f(x)]ⁿ⁺¹/(n+1) + C (n ≠ – why such patterns work and is the fallback when a pattern is not immediately recognisable.
经验丰富的学生会学习识别能通过逆向标准导数直接处理的积分。例如，∫ cos(ax+b) dx = (1/a) sin(ax+b) + C 可直接从 sin(ax+b) 的微分得出。这里使用换元 (u = ax+b) 虽然强化了模式，但在计时考试中常被略去。类似地，∫ f'(x) [f(x)]ⁿ dx = [f(x)]ⁿ⁺¹/(n+1) + C（n ≠ –1）也是一个公认的逆链式法则情形。换元法则清楚展示了这些模式为何成立，并在模式无法立即识别时作为备用方法。
9. Common Pitfalls and How to Avoid Them | 常见陷阱与规避方法
etting to replace dx fully: after writing dx = du / (du/dx), sometimes students leave an x‑ re eliminated. Another error occurs when the chosen u leads to a more complicated integral — , try an alternative u or even split the integral first. With definite integrals, mixing u‑limits and x‑limits leads to wrong answers; either change both or neither. Also, after back‑ by differentiating your answer: you should recover the original integrand. Algebraic slip‑ bstitution are common; perform each step on scrap paper before writing the final solution.
一个常见错误是忘记完全替换 dx：写出 dx = du / (du/dx) 后，有些学生仍保留着 x 项。务必确保每一个 x 和原本的 dx 都被消去。另一个错误是所选 u 导致积分反而更复杂——若情况恶化，尝试换一种 u，甚至先拆开积分。处理定积分时，混用 u 上下限和 x 上下限会得出错误答案；要么两者都变，要么都不变。此外，回代后一定要通过求导检验：应当重新得到原始被积函数。换元后展开或化简过程中的代数失误也很常见；在写下最终解答前，先在草稿上逐步核对。
10. Worked Example 1 – Polynomial Substitution | 例题 1 —— 多项式换元
Evaluate ∫ 2x (x²+3)⁴ dx.
Let u = x²+3, then du/dx = 2x → du = 2x dx.
The integral becomes ∫ u⁴ du = u⁵/5 + C.
Substituting back: (x²+3)⁵/5 + C.
You can verify by differentiating: d/dx [(x²+3)⁵/5] = 5(x²+3)⁴・2x /5 = 2x (x²+3)⁴, which matches.
This example illustrates the perfect match of substituting the inner function when its derivative appears as a factor.
计算 ∫ 2x (x²+3)⁴ dx。
令 u = x²+3，则 du/dx = 2x → du = 2x dx。
积分化为 ∫ u⁴ du = u⁵/5 + C。
回代得到：(x²+3)⁵/5 + C。
可通过求导验证：d/dx [(x²+3)⁵/5] = 5(x²+3)⁴・2x /5 = 2x (x²+3)⁴，吻合。
本例完美展示了当内层函数的导数作为因子出现时换元的匹配运用。
11. Worked Example 2 – Definite Integral with Limits | 例题 2 —— 带上下限的定积分
Evaluate ∫₀¹ 4x / √2x²+1 dx.
Let u = 2x²+1 → du/dx = 4x → du = 4x dx.
Change limits: when x=0, u=1; when x=1, u=3.
The integral becomes ∫₁³ 1/√u du = ∫₁³ u–½ du = [2u½]₁³ = 2√3 – 2√1 = 2√3 – 2.
Notice that the x‑factors were completely absorbed by du, making the u‑integral straightforward. If limits had not been changed, the back‑substitution would yield the same numeric result: 2√2x²+1 |₀¹ = 2√3 – 2, but altering limits at the start is safer under exam pressure.
计算 ∫₀¹ 4x / √2x²+1 dx。
令 u = 2x²+1 → du/dx = 4x → du = 4x dx。
变更上下限：当 x=0 时 u=1；当 x=1 时 u=3。
积分化为 ∫₁³ 1/√u du = ∫₁³ u–½ du = [2u½]₁³ = 2√3 – 2√1 = 2√3 – 2。
注意 x 因子已完全被 du 吸收，使 u 积分变得直接。若不改变上下限，回代后同样得到数值结果：2√2x²+1 |₀¹ = 2√3 – 2，但考试压力下，一开始就改变上下限更为稳妥。
12. Summary and Exam Tips | 总结与考试建议
Integration by substitution is an essential, multi‑purpose tool. Always look for composite structures first. Write the substitution clearly, compute du/dx carefully, and ensure full conversion to u. For definite integrals, convert the limits immediately. Check your answer by mental differentiation. Edexcel often embeds substitution within larger problems—such as finding areas under curves or solving differential equations—so practise recognising where a quick u‑sub can unlock a seemingly stuck integral. Keep a library of standard forms in mind, but always be ready to apply the formal substitution when the pattern is obscured.
换元积分法是一项基本且用途广泛的工具。务必首先寻找复合结构。清晰地写出代换，仔细计算 du/dx，确保完全化为 u 变量。对于定积分，立即转换积分上下限。完成答案后通过心算求导进行检验。Edexcel 考试常将换元法融入更大题目中——如求曲线下方面积或解微分方程——因此要多练习识别在何处一个快速的 u‑换元就能解开看似难办的积分。熟记标准形式库，同时时时准备在模式不明显时正式应用换元法。
Published by TutorHao | Pure 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