📚 Integration by Substitution | 换元积分法
Integration by substitution is the reverse of the chain rule and one of the most powerful techniques in the Edexcel A-Level Mathematics syllabus. It allows you to transform a complicated integral into a standard form by introducing a new variable u, often simplifying the integrand dramatically. Mastering this method is essential for tackling a wide range of functions, from polynomials and trigonometric expressions to exponentials and rational functions.
换元积分法是链式法则的逆运算，也是 Edexcel A-Level 数学大纲中最强大的积分技巧之一。通过引入新变量 u，你可以将复杂的被积函数转化为标准形式，从而大幅简化积分过程。掌握这一方法对于处理多项式、三角函数、指数函数直至有理函数等各种题型都至关重要。
1. Introduction to Integration by Substitution | 换元积分法简介
When we encounter an integral of the form ∫ f(g(x)) g' ion is often impossible. The substitution method works by letting u = g(x), so that du = g' is to recognise the inner function and its derivative appearing together in the integrand.
当我们遇到 ∫ f(g(x)) g'(x) dx 形式的积分时，直接积分往往难以进行。换元法通过令 u = g(x) 来工作，此时 du = g'(x) dx，积分就转化为 ∫ f(u) du，可能是一个可以立即计算的基本积分。关键在于识别出被积函数中同时出现了内层函数及其导数。
2. The Chain Rule in Reverse | 链式法则的逆用
Recall the chain rule: d/dx [F(g(x))] = F'(g(x)) g'(x). Integrating both sides gives ∫ F'(g(x)) g'(x) dx = F(g(x)) + C. If we set u = g(x), then du = g'(x) dx, and F'(g(x)) becomes F'(u). Thus ∫ F'(u) du = F(u) + C. This direct connection shows why substitution undoes the chain rule – we are effectively 'guessing' the antiderivative of a composite function.
回顾链式法则：d/dx [F(g(x))] = F'(g(x)) g'(x)。两边积分得到 ∫ F'(g(x)) g'(x) dx = F(g(x)) + C。如果我们设 u = g(x)，则 du = g'(x) dx，而 F'(g(x)) 就变成 F'(u)。因此 ∫ F'(u) du = F(u) + C。这种直接联系揭示了为什么换元法能够逆转链式法则——我们实际上在“猜测”复合函数的原函数。
3. Selecting the Substitution u | 选择代换变量 u
Choosing the right u is half the battle. The general strategy in Edexcel papers is to look for a function whose derivative also appears (perhaps with a constant factor) in the integrand. Priority choices include:
选对 u 是成功的一半。Edexcel 试卷中通用的策略是寻找其导数（可能相差常数倍）也出现在被积函数中的函数。优先选择包括：
The expression inside a bracket raised to a power, e.g. u = (2x+1) in ∫ (2x+1)⁵ dx.
括号内带幂次的表达式，如 ∫ (2x+1)⁵ dx 中令 u = (2x+1)。
The denominator of a rational function, e.g. u = x²+1 in ∫ 2x/(x²+1) dx.
有理函数的分母，如 ∫ 2x/(x²+1) dx 中令 u = x²+1。
a trigonometric, exponential or logarithmic function, e.g. u = sin x in ∫ cos x sin² x dx.
三角函数、指数或对数函数的参数，如 ∫ cos x sin² x dx 中的 u = sin x。
The radicand of a square root, e.g. u = x+3 in ∫ x√x+3 dx.
平方根中的根号内表达式，如 ∫ x√x+3 dx 中的 u = x+3。
Sometimes an apparently unrelated substitution simplifies the integral dramatically, but for the A-Level exam, the substitution is usually given in the question or is one of the obvious structures above.
有时表面上无关的代换也能极大简化积分，但在 A-Level 考试中，代换通常会在题目中给出，或者就是上述明显的结构之一。
4. Basic Substitution: Indefinite Integrals | 基本换元：不定积分
Example 1: Find ∫ 2x (x²+3)⁴ dx. Let u = x²+3, then du/dx = 2x ⇒ nsforms to ∫ u⁴ du = (1/5)u⁵ + C. Substituting back gives (1/5)(x²+3)⁵ + C. Notice how the '2x dx' perfectly matches du, leaving a simple power integral.
例 1：求 ∫ 2x (x²+3)⁴ dx。令 u = x²+3，则 du/dx = 2x ⇒ du = 2x dx。积分转化为 ∫ u⁴ du = (1/5)u⁵ + C。代回原变量得到 (1/5)(x²+3)⁵ + C。注意 '2x dx' 正好与 du 匹配，留下简单的幂函数积分。
Example 2: Find ∫ cos x esin x dx. Let u = sin x ⇒ du = cos x dx. The integral becomes ∫ eᵘ du = eᵘ + C = esin x + C. This highlights how substitution turns a seemingly difficult integral into one of the easiest exponential integrals.
例 2：求 ∫ cos x esin x dx。令 u = sin x ⇒ du = cos x dx。积分变为 ∫ eᵘ du = eᵘ + C = esin x + C。这个例子展示了换元如何将一个看似困难的积分变成最简单的指数积分。
5. Definite Integrals: Changing Limits | 定积分：改变积分限
limits to match the new variable. No need to rewrite in terms of the original x at the end – simply evaluate the new definite integral. The theorem states: ∫ₐᵇ f(g(x)) g'(x) dx = ∫g(a)g(b) f(u) du.
用换元法计算定积分时，必须同时改变积分上下限以匹配新变量。无需在最后置换回原变量 x——直接计算新的定积分即可。定理表述为：∫ₐᵇ f(g(x)) g'(x) dx = ∫g(a)g(b) f(u) du。
x²+1. When x=1, u=2; when x=2, u=5. Also du = 2x dx. The integral becomes ∫₂⁵ 1/u du = [ln|u| uld have to substitute back and then use the original x limits, which is more error-prone.
例：计算 ∫₁² 2x/(x²+1) dx。令 u = x²+1。当 x=1 时 u=2；x=2 时 u=5。又有 du = 2x dx。积分变为 ∫₂⁵ 1/u du = [ln|u|]₂⁵ = ln5 − ln2 = ln(5/2)。如果不改变积分限，你需要代回原变量再使用原来的 x 上下限，那样更容易出错。
6. Recognising the Derivative of a Function | 识别函数的导数
s a fraction where the numerator is exactly the derivative of the denominator. This is ∫ f'(x)/f(x) dx = ln|f(x)| e u = f(x) is a natural choice. Recognising this pattern saves time and is heavily tested.
一种特殊且频繁出现的情形是，被积函数为一个分数，分子恰好是分母的导数。此时 ∫ f'(x)/f(x) dx = ln|f(x)| + C。自然地选择 u = f(x)。识别这一模式能节省时间，也是高频考点。
Example: ∫ cos x / sin x dx = ln|sin x| e d/dx (sin x) = cos x. Compare with substitution u = sin x, du = cos x dx → ∫ 1/u du = ln|u| + C.
例：∫ cos x / sin x dx = ln|sin x| + C，因为 d/dx (sin x) = cos x。对照代换 u = sin x, du = cos x dx → ∫ 1/u du = ln|u| + C。
Another pattern: ∫ f' −1). This is just the chain rule in reverse and corresponds to the substitution u = f(x).
另一种模式：∫ f'(x) [f(x)]ⁿ dx = [f(x)]ⁿ⁺¹/(n+1) + C（n ≠ −1）。这不过是链式法则的逆用，对应于代换 u = f(x)。
7. Trigonometric Substitutions | 三角换元
ive relationships du = cos x dx, du = −sin x dx, du = sec² x dx make these very effective.
三角函数的积分经常需要诸如 u = sin x、u = cos x 或 u = tan x 的代换。导数关系 du = cos x dx、du = −sin x dx、du = sec² x dx 使得这些代换非常有效。
Example: ∫ tan x dx = ∫ (sin x)/(cos x) dx. Let u = cos x, then du = −sin x dx ⇒ −du = sin x dx. The integral becomes ∫ −1/u du = −ln|u| + C = −ln|cos x| + C = ln|sec x| + C.
例：∫ tan x dx = ∫ (sin x)/(cos x) dx。令 u = cos x，则 du = −sin x dx ⇒ −du = sin x dx。积分变为 ∫ −1/u du = −ln|u| + C = −ln|cos x| + C = ln|sec x| + C。
du = (1/2)tan² x + C. The pattern is clear once the correct inner function is identified.
∫ sec² x tan x dx：此处令 u = tan x，du = sec² x dx，得到 ∫ u du = (1/2)tan² x + C。一旦正确识别出内层函数，规律就变得清晰。
8. Exponential and Logarithmic Substitutions | 指数与对数换元
Exponential functions often simplify by letting u be the exponent. For ∫ eax+b dx, notice d/dx (ax+b) = a, so we can set u = ax+b and obtain ∫ eᵘ (1/a) du = (1/a)eax+b + C. This is a straightforward scaling.
指数函数常常通过令 u 为指数部分来简化。对于 ∫ eax+b dx，注意到 d/dx (ax+b) = a，因此设 u = ax+b 得到 ∫ eᵘ (1/a) du = (1/a)eax+b + C。这是一个直接的缩放。
Example with product: ∫ x ex² dx. Let u = x² ⇒ du = 2x dx, so (1/2)du = x dx. The integral becomes (1/2)∫ eᵘ du = (1/2)ex² + C. The 2x neatly pairs with the derivative of x².
含乘积的例子：∫ x ex² dx。令 u = x² ⇒ du = 2x dx，所以 (1/2)du = x dx。积分变为 (1/2)∫ eᵘ du = (1/2)ex² + C。2x 完美地与 x² 的导数配对。
derivative of ln x is 1/x. Set u = ln x, du = (1/x) dx. The integral becomes ∫ 1/u du = ln|u| + C = ln|ln x| + C. This recursive structure appears in Edexcel exam questions.
对数形式：∫ 1/(x ln x) dx。识别出 ln x 的导数是 1/x。令 u = ln x，du = (1/x) dx。积分变为 ∫ 1/u du = ln|u| + C = ln|ln x| + C。这种递归结构会出现在 Edexcel 考题中。
9. Substitution with Rational Functions | 有理函数换元
∫ (2x+1)/(x²+x+1) dx: here d/dx (x²+x+1) = 2x+1. Set u = x²+x+1, du = (2x+1) dx. The integral becomes ∫ 1/u du = ln|x²+x+1| + C.
当分子与分母的导数密切相关时，许多有理函数可以通过令 u 等于分母来积分。例如 ∫ (2x+1)/(x²+x+1) dx：这里 d/dx (x²+x+1) = 2x+1。设 u = x²+x+1，du = (2x+1) dx。积分变为 ∫ 1/u du = ln|x²+x+1| + C。
light adjustment with constants can help. Consider ∫ x/(x²+1) dx: let u = x²+1, du = 2x dx ⇒ (1/2)du = x dx. We get (1/2)ln|x²+1| + C. This constant-factor technique is extremely common.
即使分子不是精确的倍数，通过常数调整也能解决。考虑 ∫ x/(x²+1) dx：令 u = x²+1，du = 2x dx ⇒ (1/2)du = x dx。我们得到 (1/2)ln|x²+1| + C。这种常数因子技巧极为常见。
10. Advanced Examples with Radicals | 含根式的进阶例题
ts often require u to be the expression under the root or a linear transformation. For ∫ x√x+1 dx, let u = x+1 ⇒ x = u−1, dx = du. Then ∫ (u−1)√u du = ∫ (u3/2 − u1/2) du = (2/5)u5/2 − (2/3)u3/2 + C. Back-substitute to get the answer in x. This method – solving for x in terms of u – is useful when the derivative is not directly available.
含平方根的积分通常需要令 u 为根号内的表达式或线性变换。对于 ∫ x√x+1 dx，令 u = x+1 ⇒ x = u−1，dx = du。则 ∫ (u−1)√u du = ∫ (u3/2 − u1/2) du = (2/5)u5/2 − (2/3)u3/2 + C。回代得到以 x 表示的答案。当导数不能直接利用时，这种解出 x 用 u 表示的方法非常有用。
Another radical example: ∫ 1/(√x (1+√x)) dx. Try u = 1+√x, then du = 1/(2√x) dx ⇒ 2 du = 1/√x dx. The integral becomes ∫ (2/u) du = 2 ln|1+√x| + C. The choice of u effectively collapses the integrand.
另一个根式例题：∫ 1/(√x (1+√x)) dx。尝试 u = 1+√x，则 du = 1/(2√x) dx ⇒ 2 du = 1/√x dx。积分变为 ∫ (2/u) du = 2 ln|1+√x| + C。u 的选择有效地“折叠”了被积函数。
| Integral Form | Suggested u |
| ∫ f(ax+b) dx | u = ax+b |
| ∫ f'(x) / f(x) dx | u = f(x) |
| ∫ f'(x) [f(x)]ⁿ dx | u = f(x) |
| ∫ (g(x))ⁿ g'(x) dx | u = g(x) |
| ∫ eg(x) g'(x) dx | u = g(x) |
| ∫ sin(g(x)) g'(x) dx | u = g(x) |
11. Common Mistakes to Avoid | 常见错误
Forgetting to replace dx with du correctly. If du = g'(x) dx, then dx = du / g'(x). Always write this explicitly.
忘记正确地将 dx 替换为 du。如果 du = g'(x) dx，那么 dx = du / g'(x)。务必明确写出来。
Leaving the integral in terms of both x and u. Once u is substituted, all x's must be eliminated.
积分结果中同时出现 x 和 u。一旦代换了 u，必须消去所有的 x。
them incorrectly. Remember : lower x-limit → lower u-limit, upper x-limit → upper u-limit.
在定积分中忘记改变积分限，或者改错了。记住：x 的下限 → u 的下限，x 的上限 → u 的上限。
Dropping the constant multiple when du contains a factor like 2. Adjust by putting in the reciprocal factor outside the integral.
当 du 含有如 2 的因子时，遗漏常数倍数。通过在积分外面乘上倒数因子来调整。
Not back-substituting for indefinite integrals. Final answer must be in the original variable unless otherwise stated.
不定积分时没有回代原变量。除非题目另有说明，最终答案必须用原变量表示。
12. Summary and Key Points | 总结与要点
e inner function u = g(x) whose derivative appears (possibly up to a constant) in the integrand. (2) Compute du = g' ack to x. For definite integrals, change the limits and evaluate without back-substitution. Mastering this technique, along with pattern recognition, is essential for success in Edexcel A-Level Mathematics.
换元积分法通过巧妙变换变量，将复杂的复合形式转化为基本积分。核心步骤为：(1) 识别内层函数 u = g(x)，其导数（可能相差常数倍）出现在被积函数中。(2) 计算 du = g'(x) dx 并替换 dx。(3) 将整个积分只用 u 表示。(4) 对 u 积分。(5) 对于不定积分，代回 x；对于定积分，改变积分限并直接计算，无需回代。掌握这一技术以及模式识别能力，是 Edexcel A-Level 数学成功的关键。
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