📚 Edexcel A-Level Maths: Partial Fractions & Integration – Practice Set 137 | 爱德思A-Level数学：部分分式与积分——练习集137
This practice set from PDF Joiner (4), page 137, focuses on one of the highest-yield topics in the Edexcel A-Level Mathematics specification: decomposing rational functions into partial fractions and then integrating the result. The techniques covered here appear regularly in Pure 3 and Pure 4 exam papers, and they are usually worth between 6 and 11 marks in a single question. Mastery of the topic therefore has a direct and measurable effect on the final grade.
本练习集来自 PDF Joiner (4) 第137页，重点复习爱德思A-Level数学大纲中得分率较高的一个主题：把有理函数分解为部分分式，然后对结果进行积分。这里涉及的技巧在 Pure 3 和 Pure 4 试卷中高频出现，单题分值通常在 6 到 11 分之间。因此，掌握该主题对最终成绩有直接而可衡量的影响。
The set is designed to be worked through in order. Each section below revises a single idea, shows the standard method the examiner expects, and then records the mistake that most often costs marks. Worked examples are given in full so that every algebraic step can be checked independently.
本练习集建议按顺序完成。下面每一节复习一个独立知识点，展示考官期望看到的标准方法，然后指出最容易导致失分的错误。完整例题会逐步给出，方便读者独立核对每一步代数运算。
1. Overview of Partial Fractions and Integration | 部分分式与积分概述
Partial fraction decomposition is a technique that reverses the addition of rational expressions. It allows a single rational function to be written as a sum of simpler fractions whose denominators are the factors of the original denominator. Instead of facing one complicated quotient, you face two or three elementary pieces that can each be dealt with separately.
部分分式分解是一种逆转有理式加法运算的技巧。它能把一个有理函数写成若干个较简单分式的和，这些分式的分母是原分母的因式。于是你不再面对一个复杂的商式，而是面对两三个各自独立、可以分别处理的基本片段。
Once the expression is decomposed, each term can be integrated using standard results such as the natural logarithm or the inverse tangent. This is particularly useful because most rational functions cannot be integrated directly by inspection. A direct substitution rarely removes the difficulty: the numerator and the denominator are usually unrelated, so no single u-substitution simplifies both at once.
一旦完成分解，每一项都可以使用标准结果来积分，例如自然对数或反正切。这特别有用，因为大多数有理函数无法通过直接观察积出结果。直接换元很难消除困难：分子与分母通常没有关联，没有任何一个换元能同时简化两者。
∫ 1/(a x + b) dx = (1/a) ln|a x + b| + C
This standard integral is the main tool after decomposition, and it should be memorised carefully. The modulus sign is essential, because a x + b may be negative over part of the domain, and the logarithm of a negative number is not defined in real analysis.
这个标准积分是分解之后的主要工具，应当牢牢记住。绝对值符号必不可少，因为 a x + b 在部分定义域上可能取负值，而负数的对数在实数范围内没有定义。
The second key result used throughout this set involves a quadratic denominator that cannot be factorised further:
本练习集贯穿使用的第二个关键结论涉及无法继续分解的二次分母：
∫ 1/(x² + a²) dx = (1/a) arctan(x/a) + C
Between these two results, almost every partial fraction integration question in Pure 3 and Pure 4 can be completed. The remaining work is algebraic bookkeeping: finding the constants A, B and C correctly and completely.
有了这两个结论，Pure 3 和 Pure 4 中几乎所有的部分分式积分题都能完成。剩下的工作是代数上的整理：正确且完整地求出常数 A、B 和 C。
2. When to Use Partial Fraction Decomposition | 何时使用部分分式分解
Use partial fractions when the integrand is a rational function N(x)/D(x) whose denominator can be factorised into linear or irreducible quadratic factors. A simple substitution often does not work in these cases, so decomposition is the most reliable method. If the denominator already appears, up to a constant multiple, as the derivative of something in the numerator, then substitution is faster; otherwise partial fractions is the default route.
当被积函数是有理函数 N(x)/D(x)，且分母可以分解为线性因式或不可约二次因式时，使用部分分式。此时简单换元往往无效，因此分解是最可靠的方法。如果分母（差一个常数倍）恰好是分子中某个表达式的一阶导数，那么换元更快；否则部分分式就是默认路径。
Before starting, always check the degree of the numerator and the degree of the denominator. The method only applies directly to proper fractions. A proper fraction is one in which the degree of the numerator is strictly less than the degree of the denominator.
开始之前，务必检查分子和分母的次数。该方法仅直接适用于真分式。真分式是指分子次数严格小于分母次数的分式。
| Condition | Action |
| Degree of numerator < degree of denominator | The fraction is proper. Decompose immediately. |
| Degree of numerator = degree of denominator | Divide first. The remainder is a constant. |
| Degree of numerator > degree of denominator | Polynomial division first; the quotient may be quadratic or higher. |
| Denominator has an irreducible quadratic factor | Use (B x + C) on that factor, not a single constant. |
| Denominator has a repeated factor (x + k)ⁿ | Write n separate terms with powers 1 to n. |
The table is worth copying into your notes. In the examination, misreading the degree condition is the single most common reason for losing the first method mark before any integration has begun.
这张表值得抄进笔记。在考试中，误判次数条件是积分尚未开始就丢掉第一个方法分的最常见原因。
3. The Four Standard Denominator Cases | 四种标准分母情形
Every partial fraction question in the Edexcel specification reduces to one of four denominator structures. Recognising the structure in the first ten seconds of reading the question determines the whole shape of your answer, so it is worth learning the four cases as patterns rather than as separate techniques.
爱德思大纲中的每一道部分分式题，最终都归结为四种分母结构之一。在读到题目的前十秒内识别出结构，就决定了整个解答的形态，因此值得把四种情形当作模式来记，而不是当作互不相干的方法。
- Case 1: distinct linear factors, for example (x - 1)(x + 2).
- Case 2: a repeated linear factor, for example (x + 1)² or (x - 3)³.
- Case 3: an irreducible quadratic factor, for example x² + 4 or x² + x + 1.
- Case 4: an improper fraction, where division is required before decomposition.
- 情形一：互不相同的线性因式，例如 (x - 1)(x + 2)。
- 情形二：重复线性因式，例如 (x + 1)² 或 (x - 3)³。
- 情形三：不可约二次因式，例如 x² + 4 或 x² + x + 1。
- 情形四：假分式，需要先做除法再分解。
16, so it cannot be written as a product of two real linear factors, and it must be left intact in the decomposition.
不可约的意思是该二次式没有实根，即判别式 b² - 4a c 为负。例如 x² + 4 的判别式为 0 - 16 = -16，无法写成两个实线性因式之积，因此在分解中必须原样保留。
4. Case 1: Distinct Linear Factors | 情形一：互不相同的线性因式
For distinct linear factors, each factor receives exactly one constant numerator. The template is written before any algebra is attempted, because writing the correct template is itself a mark in most mark schemes.
对于互不相同的线性因式，每个因式对应一个常数分子。在开始任何代数运算之前先写出模板，因为在多数评分标准中，写出正确模板本身就是一个得分点。
N(x)/((x - p)(x - q)) = A/(x - p) + B/(x - q)
Multiply both sides by the original denominator to clear all fractions. The result is an identity that must hold for every value of x, which is the key that unlocks the constants.
两边同时乘以原分母，消去所有分式。得到的是一个对任意 x 都成立的恒等式，而正是这一点让我们能够求出常数。
N(x) = A(x - q) + B(x - p)
wo values are exactly the roots of the factors, which is why this shortcut is often called the cover-up rule.
最快的路径是代入。取 x = p 会使含 B 的项消失，立刻得到 A；取 x = q 则同样得到 B。这两个值恰好是因式的根，因此这个捷径常被称为遮盖法。
ives -9 = -3B, so B = 3. A quick check by expanding confirms 2(x + 2) + 3(x - 1) = 5x + 1.
以 (5x + 1)/((x - 1)(x + 2)) 的分解为例。代入 x = 1 得 6 = 3A，所以 A = 2；代入 x = -2 得 -9 = -3B，所以 B = 3。展开检验可得 2(x + 2) + 3(x - 1) = 5x + 1，结果正确。
t use, for example x = 0. This takes ten seconds and catches almost every arithmetic slip.
务必再用一个未使用过的 x 值（例如 x = 0）检验一次。这一步只要十秒，却能抓住几乎所有算术失误。
5. Case 2: Repeated Linear Factors | 情形二：重复线性因式
A repeated factor such as (x + 1)² requires two separate terms, not one. The rule is that a factor raised to the power n contributes n terms, with denominators of power 1, 2, up to n. Writing only one term is a guaranteed loss of marks and produces a decomposition that cannot be correct in general.
像 (x + 1)² 这样的重复因式需要两个独立项，而不是一个。规则是：n 次幂的因式贡献 n 个项，分母的幂次从 1 到 n。只写一项必然失分，而且得到的分解在一般情况下不可能成立。
N(x)/(x + 1)² = A/(x + 1) + B/(x + 1)²
ause when A is isolated the remaining expression still contains a factor of (x + 1), so substituting x = -1 makes it undefined rather than zero.
常数 B 仍可用遮盖法求出：两边乘以 (x + 1)² 后代入 x = -1 即可。但 A 不能用这种方法求，因为分离出 A 后剩下的表达式仍含有 (x + 1) 这个因式，代入 x = -1 会使其无定义，而不是变为零。
are accepted; coefficient comparison is usually faster and less error-prone when the numerator is linear.
求 A 的方法是：比较两边 x 的系数，或代入任意方便的值（例如 x = 0）。两种方法都被接受；当分子是一次式时，比较系数通常更快、更不易出错。
= 2. Comparing coefficients of x gives A = 3. The decomposition is 3/(x + 1) + 2/(x + 1)².
例题：分解 (3x + 5)/(x + 1)²。两边乘以分母得 3x + 5 = A(x + 1) + B。令 x = -1 得 B = 2；比较 x 的系数得 A = 3。因此分解结果为 3/(x + 1) + 2/(x + 1)²。
not the logarithm rule. This distinction is important and is discussed again in section 9.
第二项的积分使用 n = -2 的幂函数法则，而不是对数法则。这一区别很重要，第 9 节会再次讨论。
6. Case 3: Irreducible Quadratic Factors | 情形三：不可约二次因式
When a quadratic factor cannot be factorised over the reals, its numerator must be a general linear expression B x + C rather than a single constant. Using a constant here is one of the most frequently penalised errors in Pure 3, because the resulting identity is simply not solvable in general.
当二次因式在实数范围内无法分解时，它的分子必须是一般一次式 B x + C，而不能是单个常数。在这里使用常数是 Pure 3 中最常被扣分的错误之一，因为由此得到的恒等式在一般情况下根本无解。
N(x)/((x - p)(x² + a²)) = A/(x - p) + (B x + C)/(x² + a²)
l. Then B and C are found by comparing the coefficients of x² and of the constant term, or by substituting two other convenient values of x. Two equations are needed for two unknowns.
照常代入 x = p 可求出 A。接着通过比较 x² 的系数与常数项，或再代入两个方便的 x 值，可以求出 B 和 C。两个未知数需要两个方程。
After integration, the quadratic term normally splits into two pieces: a logarithmic piece from the part of the numerator proportional to the derivative of the quadratic, and an arctangent piece from what is left over. Completing the square may be required before the arctangent standard result can be applied.
积分之后，二次项通常会拆成两部分：一部分来自分子中与二次式导数成比例的部分，给出对数项；剩下的部分给出反正切项。在套用反正切标准结果之前，可能需要先配方。
the denominator can be written with a single squared bracket plus a positive constant. That form matches the standard result directly.
例如 x² + 4x + 13 = (x + 2)² + 9，因此分母可以写成一个平方括号加一个正常数。这个形式可以直接匹配标准结果。
7. Case 4: Improper Fractions and Polynomial Division | 情形四：假分式与多项式除法
If the degree of the numerator is greater than or equal to the degree of the denominator, the fraction is improper and must be divided first. Decomposing an improper fraction directly almost always produces a contradiction, because the identity cannot hold for large values of x.
如果分子次数大于或等于分母次数，该分式为假分式，必须先做除法。直接分解假分式几乎总会产生矛盾，因为该恒等式在 x 取很大值时不可能成立。
A useful intuition is to look at what happens as x becomes very large. A proper fraction tends to zero, while the quotient of two polynomials of equal degree tends to a constant, and the quotient of a higher-degree numerator tends to infinity. Partial fractions can only reproduce the decaying behaviour of a proper fraction.
一个有用的直观判断是观察 x 趋于很大时的情况。真分式趋于零，而分子分母同次的分式趋于一个常数，分子次数更高的分式则趋于无穷。部分分式只能重现真分式的衰减行为。
Worked example: integrate (x² + 1)/(x - 1). Dividing gives a quotient of x + 1 and a remainder of 2, so the fraction equals x + 1 + 2/(x - 1). Note that the division step is not optional: it changes the answer from an impossible logarithm of a quadratic to a straightforward sum of three integrable pieces.
例题：积分 (x² + 1)/(x - 1)。相除得商 x + 1、余数 2，因此该分式等于 x + 1 + 2/(x - 1)。注意除法这一步不可省略：它把无法直接积分的结果变成三块可以分别积分的简单项。
∫ (x² + 1)/(x - 1) dx = x²/2 + x + 2 ln|x - 1| + C
In the examination, show the division clearly. Some mark schemes award a mark for the quotient and a separate mark for the remainder term, so an answer written straight into the integral without showing the division may lose method marks even if the final line happens to be correct.
考试中要把除法过程写清楚。有些评分标准对商和余项分别给分，所以如果不展示除法而直接把结果写进积分式，即使最后一行碰巧正确，也可能丢掉方法分。
8. Evaluating the Constants: Cover-Up and Comparing Coefficients | 求常数：遮盖法与比较系数
Two methods dominate the mark schemes: substitution, including the cover-up shortcut, and comparison of coefficients. Strong candidates choose whichever is faster for the particular structure, and sometimes use a mixture of both in the same question.
评分标准中最常见的是两种方法：代入法（含遮盖法捷径）与比较系数法。水平较高的考生会针对具体结构选择更快的一种，有时在同一题中混用两种方法。
- Cover-up works best when all factors are distinct and linear.
- Cover-up also gives the highest-power term when a factor is repeated.
- Coefficient comparison is necessary for irreducible quadratic numerators B x + C.
- uting a simple value such as x = 0 is a reliable third route when the algebra looks messy.
- Always finish with a numerical check at a value of x you have not used.
- 当所有因式互不相同且为线性时，遮盖法最快。
- 当某个因式重复时，遮盖法可以求出幂次最高的那一项。
- 当不可约二次因式的分子形如 B x + C 时，必须使用比较系数法。
- 当代数式看起来很乱时，代入 x = 0 这样简单的值是可靠的第三条路径。
- 最后一定要用一个未使用过的 x 值做数值检验。
to work with than the original quotient, and the examiner can follow the logic at a glance.
每一步都要写成清晰的恒等式，而不是两边都带分式的方程。恒等式 5x + 1 = A(x + 2) + B(x - 1) 远比原始商式好处理，考官也能一眼看懂逻辑。
s unless the question requires it. Exact fractions such as A = 2/3 are preferred, and in a "show that" question a decimal answer will frequently fail to match the printed result.
除非题目要求，不要把 A 和 B 写成小数。像 A = 2/3 这样的精确分数更受青睐，而在证明题中，小数答案往往无法与题面给出的结果吻合。
9. Integration After Decomposition: ln and arctan Results | 分解后的积分：对数与反正切结果
After decomposition, each term falls into one of three integration patterns. Identifying the pattern correctly is the final skill tested by this topic, and it is where many otherwise correct solutions lose their last few marks.
分解之后，每一项都落入三种积分模式之一。正确识别模式是本主题考查的最后一项技能，也是许多本来正确的解答丢掉最后几分的地方。
| Integrand | Standard Result |
| 1/(a x + b) | (1/a) ln|a x + b| + C |
| 1/(a x + b)ⁿ, n ≠ 1 | (1/a) × (a x + b)¹⁻ⁿ / (1 - n) + C |
| 1/(x² + a²) | (1/a) arctan(x/a) + C |
| f'(x)/f(x) | ln|f(x)| + C |
The third row is the one that catches students who have memorised the logarithm rule too enthusiastically. The integral of 1/(x + 1)² is not a logarithm; it is minus one over x + 1. Applying the logarithm rule to a repeated linear factor produces a right-looking answer that is completely wrong.
第三行最容易抓住那些过度套用对数规则的学生。1/(x + 1)² 的积分不是对数，而是负的 1/(x + 1)。把对数规则用在重复线性因式上，会得到一个看起来合理、实际完全错误的答案。
ple from section 5. Since (3x + 5)/(x + 1)² = 3/(x + 1) + 2/(x + 1)², the integral is 3 ln|x + 1| t term uses the logarithm rule with a = 1, and the second uses the power rule with n = -2.
继续第 5 节的例题。由于 (3x + 5)/(x + 1)² = 3/(x + 1) + 2/(x + 1)²，积分为 3 ln|x + 1| - 2/(x + 1) + C。第一项使用 a = 1 的对数规则，第二项使用 n = -2 的幂函数规则。
For a quadratic denominator, write the numerator as a multiple of the derivative of the denominator plus a constant remainder. This single step converts one difficult integral into a logarithm plus a standard arctangent, and it is worth practising until it becomes automatic.
对于二次分母，把分子写成分母导数的倍数加上一个常数余项。这一步就能把一个困难的
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