📚 Integration by Substitution and By Parts | 换元积分法与分部积分法
In Edexcel A-Level Maths, integration questions often go beyond standard results. Two powerful techniques are integration by substitution and integration by parts. This revision article explains when and how to use each method, with worked examples, exam tips and common mistakes.
在 Edexcel A-Level 数学中，积分题往往超出基本公式的范围。换元积分法和分部积分法是两个强大技巧。本文介绍每种方法的使用时机与步骤，包含例题、考试技巧和常见错误。
1. Recognising the Reverse Chain Rule | 识别反向链式法则
If an integrand is a composite function multiplied by the derivative of the inner function, you can integrate directly. In general, if u = g(x), then ∫ f(g(x)) g'(x) dx = ∫ f(u) du. This is the reverse of the chain rule.
如果被积函数是复合函数乘以内层函数的导数，就可以直接积分。一般地，若 u = g(x)，则 ∫ f(g(x)) g'(x) dx = ∫ f(u) du。这就是链式法则的逆运算。
For example, in ∫ cos(5x) dx, the inner function is 5x and its derivative is 5. We can adjust the constant by writing ∫ cos(5x) dx = (1/5) sin(5x) + C.
例如，在 ∫ cos(5x) dx 中，内层函数是 5x，其导数为 5。我们可以调整常数，写成 ∫ cos(5x) dx = (1/5) sin(5x) + C。
2. Integration by Substitution: Core Method | 换元积分法：核心方法
Choose u = g(x), then find du = g'(x) dx. Rewrite the whole integral in terms of u only, integrate with respect to u, and finally substitute back for x. If the integral is definite, either change the limits to u-values or rewrite the antiderivative in terms of x before evaluating.
选择 u = g(x)，然后求出 du = g'(x) dx。将整个积分仅用 u 表示，对 u 积分，最后再代回 x。若为定积分，可以将积分上下限转换为 u 值，或者先用 x 表示原函数再代入原上下限。
The goal is to convert a complicated integral into one that matches a standard form, such as ∫ uⁿ du, ∫ eᵘ du, or ∫ cos u du.
目标是把复杂积分转化为标准形式，例如 ∫ uⁿ du、∫ eᵘ du 或 ∫ cos u du。
3. Choosing the Substitution | 如何选择代换
Usually choose u as the inner part of a composite function: the expression inside brackets, under a root, in a denominator, or inside a trig or exponential function. A good substitution makes the new integral simpler than the original.
通常选择复合函数的内层作为 u：括号内、根号下、分母中，或三角/指数函数内部。一个好的代换会使新积分比原积分更简单。
For ∫ x(2x² + 1)³ dx, the natural choice is u = 2x² + 1, because the derivative du = 4x dx is closely related to the remaining x dx.
对于 ∫ x(2x² + 1)³ dx，自然选择 u = 2x² + 1，因为其导数 du = 4x dx 与剩下的 x dx 密切相关。
4. Worked Example: ∫ x(2x² + 1)³ dx | 例题：∫ x(2x² + 1)³ dx
Let u = 2x² + 1, so du = 4x dx. Therefore x dx = du/4. The integral becomes (1/4)∫ u³ du.
令 u = 2x² + 1，则 du = 4x dx。因此 x dx = du/4。积分变为 (1/4)∫ u³ du。
(1/4)∫ u³ du = (1/4)(u⁴/4) + C = (1/16)u⁴ + C
Substitute back u = 2x² + 1 to obtain (1/16)(2x² + 1)⁴ + C.
代回 u = 2x² + 1，得到 (1/16)(2x² + 1)⁴ + C。
5. Worked Example: Definite Integral with Limits | 定积分例题
Evaluate ∫ from 0 to 1 of 2x(x² + 3)² dx. Let u = x² + 3, so du = 2x dx. When x = 0, u = 3; when x = 1, u = 4.
计算 ∫₀¹ 2x(x² + 3)² dx。令 u = x² + 3，则 du = 2x dx。当 x = 0 时 u = 3；当 x = 1 时 u = 4。
The integral becomes ∫ from 3 to 4 of u² du.
积分变为 ∫₃⁴ u² du。
[u³/3] from 3 to 4 = (64/3) − (27/3) = 37/3
Always remember to change the limits when you change the variable in a definite integral.
在定积分中换元时，一定要记得同时更换积分上下限。
6. Integration by Parts: Formula | 分部积分法公式
Integration by parts reverses the product rule. If u and v are functions of x, the formula is:
分部积分法是乘积法则的逆运算。若 u 和 v 是 x 的函数，公式为：
∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx
Alternatively, in a shorter notation, ∫ u dv = uv − ∫ v du. This is useful when the product of two different types of functions appears, such as algebraic times exponential or logarithmic times algebraic.
简写形式为 ∫ u dv = uv − ∫ v du。当出现两种不同类型函数的乘积时很常用，例如代数函数乘以指数函数，或对数函数乘以代数函数。
7. LIATE Rule and Choosing u | LIATE 法则与 u 的选择
To choose u in integration by parts, the LIATE rule gives a useful priority order: Logarithm, Inverse trig, Algebraic, Trigonometric, Exponential. Choose u as the type appearing first in this list; let dv be the remaining part.
在分部积分中选择 u 时，LIATE 法则提供了优先顺序：对数函数、反三角函数、代数函数、三角函数、指数函数。选择列表中靠前的类型作为 u，其余部分作为 dv。
| Priority | Function Type | 函数类型 |
| 1 | Logarithmic | 对数函数 |
| 2 | Inverse trigonometric | 反三角函数 |
| 3 | Algebraic | 代数函数 |
| 4 | Trigonometric | 三角函数 |
| 5 | Exponential | 指数函数 |
8. Worked Example: ∫ x eˣ dx | 例题：∫ x eˣ dx
Here u = x because it is algebraic, and dv = eˣ dx because it is exponential. Then du = dx and v = eˣ.
这里 u = x，因为它是代数函数；dv = eˣ dx，因为它是指数函数。则 du = dx，v = eˣ。
∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C
Notice that the new integral is simpler than the original, which is exactly what we want.
注意新积分比原积分更简单，这正是我们想要的结果。
9. Worked Example: ∫ ln x dx | 例题：∫ ln x dx
To integrate ln x, take u = ln x and dv = dx. Then du = (1/x) dx and v = x.
对 ln x 积分时，取 u = ln x，dv = dx。则 du = (1/x) dx，v = x。
∫ ln x dx = x ln x − ∫ x(1/x) dx = x ln x − x + C
This example shows how integration by parts can handle functions that do not have a standard antiderivative immediately.
这个例子说明分部积分法可以处理没有直接标准原函数的函数。
10. Repeated Integration by Parts | 重复使用分部积分法
Some integrals require more than one application of the method. For ∫ x² eˣ dx, start with u = x² and dv = eˣ dx. After one step, the power of x decreases by one.
有些积分需要多次使用分部积分法。对于 ∫ x² eˣ dx，先令 u = x²，dv = eˣ dx。一次分部积分后，x 的幂次降低一次。
Repeat the process until the algebraic part disappears or the integral returns to a known form. Keep the same type of choice for u at each stage.
重复这一过程，直到代数部分消失或积分回到已知形式。每一步中 u 的类型选择要保持一致。
11. Common Mistakes | 常见错误
Common errors include forgetting the constant of integration C, changing definite limits incorrectly, choosing the wrong u, and losing track of signs. For definite integrals with substitution, do not substitute the original x-limits into a u-antiderivative.
常见错误包括忘记积分常数 C、更换定积分上下限时出错、选择错误的 u 以及符号混乱。对于使用换元法的定积分，不要把原来的 x 上下限代入用 u 表示的原函数。
Always differentiate your final answer to check that it returns to the original integrand.
始终将最终结果求导，检查是否回到原来的被积函数。
12. Summary and Exam Tips | 总结与考试技巧
If you see a product of two different function types, try integration by parts. If you see a composite function with its inner derivative nearby, try substitution. Show all substitution steps clearly in exams, and simplify definite limits carefully.
如果看到两种不同类型函数的乘积，优先尝试分部积分。如果看到复合函数且其内层导数就在旁边，优先尝试换元法。考试中要清晰写出所有换元步骤，并仔细化简定积分上下限。
Practise recognising patterns quickly: ∫ f(g(x)) g'(x) dx for substitution, and ∫ u dv for integration by parts. Mastery of these two techniques will unlock many high-mark Edexcel A-Level integration questions.
练习快速识别模式：∫ f(g(x)) g'(x) dx 用换元法，∫ u dv 用分部积分法。掌握这两种技巧，就能解决许多 Edexcel A-Level 高分值积分题。
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