Integration Using Partial Fractions for Edexcel A-Level Maths | 利用部分分式积分（Edexcel A-Level 数学）

📚 Integration Using Partial Fractions for Edexcel A-Level Maths | 利用部分分式积分（Edexcel A-Level 数学）

Integration of rational functions is one of the most skill-intensive topics in the Edexcel A-Level Mathematics syllabus. Very often, integrals involving algebraic fractions with polynomial denominators cannot be tackled by simple substitution or by recognising standard derivatives. Partial fractions provide a systematic method to rewrite a complicated fraction as a sum of simpler fractions, each of which can be integrated directly to give logarithms, inverse tangents or simple rational functions. This article covers every type of denominator decomposition required in the exam, from distinct linear factors to repeated and irreducible quadratic factors, along with full worked examples and common pitfalls.

有理函数的积分是爱德思A-Level数学大纲中对技巧要求最高的主题之一。很多时候，带有多项式分母的代数分式无法通过简单的换元或识别标准导数来积分。部分分式提供了一套系统的方法，能把一个复杂分式改写为一系列简单分式之和，而每一个简单分式都可以直接积出对数、反正切或简单的有理函数。本文覆盖了考试中要求的每一种分母分解类型，包括不重复的线性因子、重复线性因子和不可约二次因子，并配有完整的例题和常见误区。

1. Why Partial Fractions Are Essential for Integration | 为什么部分分式对积分至关重要

In Edexcel A-Level Pure Mathematics, integration techniques are extended well beyond reverse differentiation. When the integrand is a rational function like (3x+5)/(x² - x - 6), the degree of the numerator is lower than that of the denominator, but direct integration is not obvious. Partial fractions decompose such expressions into components whose antiderivatives are immediately known: A/(x - a) integrates to A ln|x - a|, while linear numerators over quadratics often lead to arctan forms. This process is frequently examined in questions combining algebraic manipulation with definite integration or differential equations.

在爱德思A-Level纯数学中，积分技巧远超简单的逆微分。当被积函数是一个像 (3x+5)/(x² - x - 6) 这样的有理函数时，分子次数低于分母次数，但直接积分并不显然。部分分式将这类表达式分解为若干部分，其原函数一目了然：A/(x - a) 可积出 A ln|x - a|，而二次分母上的线性分子常常导出 arctan 形式。这个过程经常在结合代数操作与定积分或微分方程的问题中出现。

2. Prerequisite Skill: Polynomial Long Division | 前置技能：多项式长除法

Before jumping into partial fractions, you must check whether the rational function is proper. A fraction is improper if the degree of the numerator is greater than or equal to the degree of the denominator. For example, (x³+2x)/(x²+1) is improper. In such cases, perform polynomial long division to express the fraction as a polynomial plus a proper fraction. The partial fraction decomposition is then applied only to the proper fraction part. Edexcel exam questions often deliberately include improper fractions to test your algebraic fluency.

在进入部分分式之前，必须先检查有理函数是否为真分式。如果分子次数大于或等于分母次数，该分式就是假分式。例如，(x³+2x)/(x²+1) 就是假分式。这时需要通过多项式长除法，将其表示为一个多项式加上一个真分式。部分分式分解仅应用于真分式部分。爱德思试卷中的题目经常故意包含假分式，以检验你的代数熟练度。

For instance, divide (x³+2x) by (x²+1) to obtain x + (x)/(x²+1). The remaining proper fraction (x)/(x²+1) is then ready for partial fractions if needed.

例如，用长除法将 (x³+2x) 除以 (x²+1)，得到 x + (x)/(x²+1)。剩余的真分式 (x)/(x²+1) 随后可根据需要进行部分分式分解。

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3. Distinct Linear Factors | 不重复的线性因子

When the denominator factorises into distinct linear factors, say (x - a)(x - b), the partial fraction decomposition takes the form:

当分母可分解为不重复的线性因子，例如 (x - a)(x - b) 时，部分分式的形式为：

1/((x - a)(x - b)) = A/(x - a) + B/(x - b)

f x. Substituting x = a eliminates B, giving A directly. This is always the quickest method.

为求出常数 A 和 B，将两边同乘以分母 (x - a)(x - b)，然后可采用比较系数法或代入方便的 x 值。代入 x = a 可直接消去 B 求出 A。这始终是最快捷的方法。

Once A and B are found, integration becomes straightforward:

一旦求得 A 和 B，积分就变得十分简单：

∫ [A/(x - a) + B/(x - b)] dx = A ln|x - a| + B ln|x - b| + C

This is the foundation for all more complex cases and appears frequently in Edexcel Paper 1.

这是处理所有更复杂情形的基础，并频繁出现在爱德思试卷1中。

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4. Repeated Linear Factors | 重复的线性因子

If the denominator contains a repeated linear factor, such as (x - a)², the decomposition must include a constant over the first power and another constant over the square:

如果分母包含重复的线性因子，比如 (x - a)²，部分分式必须包含常数除以一次幂和另一个常数除以平方项：

g(x)/((x - a)²) = A/(x - a) + B/(x - a)²

For a factor (x - a)ⁿ, you need terms A₁/(x - a) + A₂/(x - a)² + … + Aₙ/(x - a)ⁿ. Notice that the numerators are constants, not polynomials.

对于因子 (x - a)ⁿ，需要设置 A₁/(x - a) + A₂/(x - a)² + … + Aₙ/(x - a)ⁿ 各项。注意分子是常数，而不是多项式。

(x - a). Together with the logarithmic term from A/(x - a), the complete integral is straightforward once constants are evaluated.

对 B/(x - a)² 项积分只需运用简单的幂法则：∫ B(x - a)⁻² dx = -B/(x - a)。在求出所有常数后，结合来自 A/(x - a) 的对数项，整个积分便一目了然。

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5. Irreducible Quadratic Factors | 不可约的二次因子

When a denominator includes a quadratic factor that cannot be factorised into real linear factors, such as x² + 1 or x² + x + 1, the numerator in the partial fraction must be linear:

当分母包含无法分解为实线性因子的二次因子时，比如 x² + 1 或 x² + x + 1，部分分式中的分子必须为线性表达式：

g(x)/((ax² + bx + c)(...)) = (Ax + B)/(ax² + bx + c) + ...

To integrate (Ax + B)/(ax² + bx + c), complete the square in the denominator and split the numerator into a multiple of the derivative of the denominator plus a constant. The first part integrates to a natural logarithm, and the second leads to an arctangent function.

要积分 (Ax + B)/(ax² + bx + c)，可将分母配方，同时将分子拆分为分母导数的倍数加上一个常数。第一部分积出自然对数，第二部分则引出反正切函数。

This technique is highly examinable because it blends algebraic decomposition with trigonometric integration, and students often lose marks by misidentifying the correct form or by errors when completing the square.

这种技巧在考试中很常见，因为它将代数分解与三角函数积分结合在一起，学生常因错误识别正确形式或在配平方时出错而失分。

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6. Step-by-Step Procedure for Integration | 积分的完整步骤

The systematic approach for integrating rational functions using partial fractions can be summarised in a clear sequence. Following this sequence helps avoid careless slips under exam pressure.

利用部分分式积分有理函数的系统方法可以总结为一个清晰的流程。在考试压力下遵循这个流程有助于避免粗心失误。

Step 1: Ensure the fraction is proper. If not, perform polynomial division.
Step 2: Factorise the denominator completely, including complex conjugate quadratic factors if needed.
Step 3: Write down the general form of the partial fractions according to the factor types.
Step 4: Clear the denominator and solve for the unknown constants using substitution or equating coefficients.
Step 5: Rewrite the original integrand as the sum of the partial fractions.
Step 6: Integrate each term individually, using ln, power rules, or completing the square as appropriate.
Step 7: Combine the results and add the constant of integration. If evaluating a definite integral, apply limits carefully.

第一步： 确保分式为真分式，若不是，则执行多项式除法。
第二步： 将分母完全分解因式，必要时包括共轭复二次因子。
第三步： 根据因子类型写出部分分式的一般形式。
第四步： 清除分母，并通过代入数值或比较系数解出未知常数。
第五步： 将原被积函数改写为部分分式之和。
第六步： 逐项积分，酌情使用 ln、幂次法则或配平方。
第七步： 合并结果并加上积分常数。如果计算定积分，小心代入上下限。

This methodology is directly aligned with the mark schemes of Edexcel 9MA0 papers, so internalising it is a sure way to secure full marks on integration problems.

这一方法与爱德思 9MA0 试卷的评分标准完全一致，因此内化这套方法能够确保在积分题上获得满分。

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7. Worked Example 1: Two Distinct Linear Factors | 例题1：两个不重复的线性因子

x - 2) dx. This is a typical Edexcel question where the denominator factorises neatly and the partial fractions lead to logarithmic answers.

考虑积分 ∫ (2x + 1)/(x² - x - 2) dx。这是一道典型的爱德思考题，分母可清晰地分解，部分分式导出对数函数答案。

Factorise the denominator: x² - x - 2 = (x - 2)(x + 1).

将分母因式分解：x² - x - 2 = (x - 2)(x + 1)。

Write the partial fraction decomposition:

写出部分分式分解形式：

(2x + 1)/((x - 2)(x + 1)) = A/(x - 2) + B/(x + 1)

1) + B(x - 2). Substitute x = 2: 5 = 3A → A = 5/3. Substitute x = -1: -1 = -3B → B = 1/3.

两边同乘 (x - 2)(x + 1)：2x + 1 = A(x + 1) + B(x - 2)。代入 x = 2：5 = 3A → A = 5/3。代入 x = -1：-1 = -3B → B = 1/3。

The integrand becomes (5/3)/(x - 2) + (1/3)/(x + 1). Integrate:

被积函数变为 (5/3)/(x - 2) + (1/3)/(x + 1)。积分：

∫ [ (5/3)/(x - 2) + (1/3)/(x + 1) ] dx = (5/3) ln|x - 2| + (1/3) ln|x + 1| + C

This example illustrates every step and shows that careful evaluation of constants is vital; a slip in 5/3 or 1/3 loses method marks in the exam.

这个例子展示了整个过程，并表明准确求出常数至关重要；若将 5/3 或 1/3 算错，会在考试中丢失方法分。

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8. Worked Example 2: Repeated Linear Denominator | 例题2：重复的线性分母

Evaluate ∫ (3x + 5)/(x + 1)² dx.

计算 ∫ (3x + 5)/(x + 1)² dx。

Here the denominator is a single repeated linear factor, so the form is:

这里分母为单个重复的线性因子，其形式为：

(3x + 5)/((x + 1)²) = A/(x + 1) + B/(x + 1)²

+ A + B. Equate coefficients: for x, A = 3; for constants, A + B = 5 → 3 + B = 5 → B = 2.

乘开：3x + 5 = A(x + 1) + B。展开：3x + 5 = Ax + A + B。比较系数：x 项系数 A = 3；常数项 A + B =

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