📚 Edexcel A-Level Pure Maths: Integration by Substitution and Partial Fractions | 爱德思A-Level纯数：换元积分与部分分式
In Edexcel A-Level Pure Mathematics, integration questions often look more difficult than the standard list of results. Many integrals from past papers require you to simplify, split or substitute before you can apply a known rule. This article focuses on two key tools: partial fractions and integration by substitution. Master these ideas and you will be able to turn complicated rational or composite functions into standard integrals.
在爱德思 A-Level 纯数考试中，积分题常常比标准公式表看起来更难。往年真题中的许多积分都要求你先化简、拆分或换元，然后才能套用已知规则。本文重点讲解两大核心工具：部分分式法和换元积分法。掌握这些方法后，你就能把复杂的有理函数或复合函数转化为标准积分。
1. Why Integration Techniques Matter | 为什么积分技巧如此重要
A direct integration formula only works when the integrand is already in a recognised form, such as xⁿ, eˣ, sin x or 1/(ax+b). In an Edexcel exam, however, the function is often built from two or more pieces: a rational expression with a product of linear factors, or a product like x√(x²+1) where one part is nearly the derivative of another. Techniques such as partial fractions and substitution allow you to rewrite the problem so that a standard result can be used.
直接积分公式只在被积函数已经是标准形式时有效，例如 xⁿ、eˣ、sin x 或 1/(ax+b)。但在爱德思考试中，函数常常由多个部分构成：比如分母为线性因式乘积的有理式，或者像 x√(x²+1) 这样的乘积，其中一部分接近另一部分的导数。部分分式和换元法能帮助你重新整理问题，从而套用标准结果。
These techniques are not just isolated tricks; they are expected at A-Level, especially in Pure Mathematics 3 and 4. The examiner wants to see whether you can recognise structure, choose a sensible method, and carry out algebraic manipulation accurately before integrating.
这些技巧并非孤立的解题窍门，而是 A-Level 考试明确要求的能力，尤其在纯数第三、第四单元中。考官希望看到你能否识别结构、选择合理方法，并在积分之前准确完成代数变形。
2. Core Idea: Rewriting Before Integrating | 核心思想：先变形再积分
Many Edexcel questions begin with a rational function such as (2x + 1)/(x² + x – 6). The denominator factorises as (x + 3)(x – 2), but the integrand is not a simple reciprocal of a linear term. If you can split it into A/(x + 3) + B/(x – 2), then each piece integrates to a natural logarithm. Partial fractions are exactly this splitting process.
许多爱德思考题会先给出一个有理函数，例如 (2x + 1)/(x² + x – 6)。分母可以因式分解为 (x + 3)(x – 2)，但被积函数并不是简单的线性项倒数。如果你能把它拆成 A/(x + 3) + B/(x – 2)，那么每一项都能积成自然对数。部分分式法正是这种拆分过程。
Similarly, substitution rewrites the integrand by introducing a new variable u. If one factor is the derivative of an inner function, the chain rule can be used in reverse. For example, in ∫ x√(x²+1) dx, the factor x is closely related to the derivative of x²+1, so the substitution u = x²+1 turns the integrand into a simple power of u.
同样，换元法通过引入新变量 u 来重写被积函数。如果其中一个因子是内层函数的导数，就可以逆向使用链式法则。例如在 ∫ x√(x²+1) dx 中，因子 x 与 x²+1 的导数密切相关，因此令 u = x²+1 就能把被积函数变成 u 的简单幂函数。
3. Partial Fractions with Linear Denominators | 线性分母的部分分式
For a rational function p(x)/q(x) where the degree of p(x) is lower than the degree of q(x) and q(x) has distinct linear factors, start by writing the fraction as a sum of unknown constants over each factor. For example, (2x+1)/((x+3)(x-2)) = A/(x+3) + B/(x-2).
对于有理函数 p(x)/q(x)，如果分子 p(x) 的次数低于分母 q(x) 的次数，且 q(x) 含有不同的线性因式，首先要把它写成每个因式上带未知常数的和。例如 (2x+1)/((x+3)(x-2)) = A/(x+3) + B/(x-2)。
Multiply through by the full denominator to obtain the identity:
两边同时乘以完整分母，得到恒等式：
2x + 1 ≡ A(x – 2) + B(x + 3)
You can find A and B by substituting convenient x-values. Setting x = 2 removes the A term, giving 5 = 5B, so B = 1. Setting x = -3 removes the B term, giving -5 = -5A, so A = 1.
你可以通过代入方便的 x 值来求出 A 和 B。令 x = 2 可以消去含 A 的项，得到 5 = 5B，所以 B = 1。令 x = -3 可以消去含 B 的项，得到 -5 = -5A，所以 A = 1。
Therefore the integrand becomes 1/(x+3) + 1/(x-2), and the integral is ln|x+3| + ln|x-2| + c.
因此被积函数变为 1/(x+3) + 1/(x-2)，积分结果为 ln|x+3| + ln|x-2| + c。
4. Handling Repeated Linear Factors | 处理重复线性因式
If the denominator contains a repeated factor such as (x-1)², the partial fraction form must include all powers up to that repeated power. For example, 1/((x-1)²(x+2)) should be written as A/(x-1) + B/(x-1)² + C/(x+2).
如果分母含有重复因式，例如 (x-1)²，部分分式的形式必须包含直到该重复次数的所有幂次。例如 1/((x-1)²(x+2)) 应写成 A/(x-1) + B/(x-1)² + C/(x+2)。
This matters because each term integrates differently. The term A/(x-1) gives A ln|x-1|, while B/(x-1)² integrates to -B/(x-1). Do not simply write A/(x-1) + B/(x+2), as that would miss the (-2) power term and produce an incorrect identity.
这一点很重要，因为各项的积分方式不同。A/(x-1) 积得 A ln|x-1|，而 B/(x-1)² 积得 -B/(x-1)。不要只写成 A/(x-1) + B/(x+2)，那样会漏掉负二次幂项，导致恒等式不成立。
5. Quadratic Denominator Forms | 二次分母的形式
When a denominator factor is a quadratic that cannot be factorised over real numbers, such as x²+1, the partial fraction numerator must be linear. The general form is (Ax+B)/(x²+1). For example, (3x+2)/((x+1)(x²+1)) becomes A/(x+1) + (Bx+C)/(x²+1).
当分母中的某个因式是无法在实数范围内分解的二次式，例如 x²+1，部分分式的分子必须是一次式。一般形式为 (Ax+B)/(x²+1)。例如 (3x+2)/((x+1)(x²+1)) 可分解为 A/(x+1) + (Bx+C)/(x²+1)。
After finding the constants, the integral may split into two known types. A term like 2x/(x²+1) integrates to ln|x²+1|, and a term like 1/(x²+1) integrates to arctan x. This is a standard Edexcel result, so be ready to recognise it without an extra substitution.
求出常数后，积分可能分成两类已知结果。像 2x/(x²+1) 这样的项积分为 ln|x²+1|，而像 1/(x²+1) 这样的项积分为 arctan x。这是爱德思考试中的标准结果，因此要能够不借助额外换元就识别出来。
6. Integration by Substitution: The Chain Rule in Reverse | 换元积分：链式法则的逆用
Integration by substitution is the reverse of the chain rule. If an integrand can be written as f(g(x))g'(x), then the substitution u = g(x) gives du/dx = g'(x), so du = g'(x) dx. The integral becomes ∫ f(u) du, which is usually easier.
换元积分法是链式法则的逆运算。如果被积函数可以写成 f(g(x))g'(x)，那么令 u = g(x)，得到 du/dx = g'(x)，即 du = g'(x) dx。积分就变成 ∫ f(u) du，通常会简单很多。
For example, consider ∫ x√(x²+1) dx. Let u = x²+1, so du/dx = 2x, and therefore x dx = (1/2) du. The integral becomes ∫ (1/2)√u du, which is (1/2) × (2/3)u^(3/2) + c = (1/3)(x²+1)^(3/2) + c.
例如，考虑 ∫ x√(x²+1) dx。令 u = x²+1，则 du/dx = 2x，所以 x dx = (1/2) du。积分变为 ∫ (1/2)√u du，结果为 (1/2) × (2/3)u^(3/2) + c = (1/3)(x²+1)^(3/2) + c。
7. Choosing the Substitution u | 如何选择换元变量 u
A good choice for u is usually the inner function of a composite expression, especially when its derivative also appears in the integrand. Common choices include u = ax+b, u = x²+1, u = sin x, u = cos x, u = ln x or u = eˣ.
选择 u 的好方法通常是取复合表达式的内层函数，尤其是当它的导数也出现在被积函数中时。常见的选择包括 u = ax+b、u = x²+1、u = sin x、u = cos x、u = ln x 或 u = eˣ。
For example, in ∫ sin³x cos x dx, take u = sin x. Then du = cos x dx, so the integral becomes ∫ u³ du = (1/4)u⁴ + c = (1/4)sin⁴x + c. The key is to check that after substitution, no original variable remains.
例如，在 ∫ sin³x cos x dx 中，令 u = sin x。则 du = cos x dx，积分变为 ∫ u³ du = (1/4)u⁴ + c = (1/4)sin⁴x + c。关键在于检查换元后不能再有原来的变量残留。
8. Definite Integrals: Changing Limits | 定积分：换限
When you use substitution in a definite integral, you can either change the limits and evaluate in terms of u, or integrate back to x before substituting the original limits. Changing the limits is usually cleaner. If u = g(x), the lower and upper limits become g(a) and g(b).
在定积分中使用换元法时，你可以先改变积分限并用 u 求值，也可以先换回 x
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