Partial Fractions: Splitting and Simplifying | 部分分数的拆分与化简

📚 Partial Fractions: Splitting and Simplifying | 部分分数的拆分与化简

Partial fractions is a powerful algebraic technique used to break a single complicated rational expression into a sum of simpler fractions. This method is indispensable in A-Level Mathematics, particularly when integrating rational functions or expanding them in binomial series.

部分分数是一种强大的代数技巧，用于将一个复杂的有理表达式拆分成若干个更简单分数之和。在A-Level数学中，这种方法不可或缺，尤其是在对有理函数进行积分或将其进行二项式展开时。

1. What Are Partial Fractions? | 什么是部分分数？

A rational expression is a fraction where both the numerator and denominator are polynomials. Partial fractions involve expressing such a fraction as a sum of simpler fractions whose denominators are the factors of the original denominator.

有理表达式是指分子和分母都是多项式的分数。部分分数法是将这样一个分数表示成若干个更简单分数的和，这些简单分数的分母是原分母的因子。

For example, the expression \(\frac{3x+1}{(x-1)(x+2)}\) can be decomposed as:

例如，表达式 \(\frac{3x+1}{(x-1)(x+2)}\) 可以分解为：

\(\frac{3x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}\)

where A and B are constants to be determined. This decomposition allows us to work with much simpler pieces.

其中 A 和 B 是待定常数。这种分解使我们能够处理更简单的部分。

2. Case 1: Distinct Linear Factors | 情形一：不同线性因子

When the denominator consists of distinct linear factors — that is, factors of the form (ax + b) where no two factors are the same — the partial fraction decomposition takes the simplest form.

当分母由互不相同的线性因子组成时（即形如 (ax + b) 的因子，且任意两个因子都不相同），部分分数分解采用最简单的形式。

Consider the fraction \(\frac{5x-2}{(x-1)(x-3)}\). We write:

考虑分数 \(\frac{5x-2}{(x-1)(x-3)}\)。我们写作：

\(\frac{5x-2}{(x-1)(x-3)} = \frac{A}{x-1} + \frac{B}{x-3}\)

Multiplying both sides by (x−1)(x−3) gives:

两边同时乘以 (x−1)(x−3) 得：

\(5x−2 = A(x−3) + B(x−1)\)

To find A, substitute x = 1 (which eliminates B): A(1−3) = 5(1)−2, so A = −3⁄2.

要求 A，代入 x = 1（这样消去 B）：A(1−3) = 5(1)−2，所以 A = −3⁄2。

To find B, substitute x = 3: B(3−1) = 5(3)−2, so B = 13⁄2.

要求 B，代入 x = 3：B(3−1) = 5(3)−2，所以 B = 13⁄2。

\(\frac{5x-2}{(x-1)(x-3)} = \frac{-3}{2(x-1)} + \frac{13}{2(x-3)}\)

This “substitution” technique is quick and reliable when all factors are distinct linear factors.

当所有因子均为不同线性因子时，“代入法”快速且可靠。

3. Case 2: Repeated Linear Factors | 情形二：重复线性因子

When the denominator contains a repeated linear factor, such as (x − 2)², the decomposition must include terms for each power of that factor.

当分母含有重复线性因子（如 (x − 2)²）时，分解必须包含该因子的每一幂次项。

For example:

例如：

\(\frac{3x^{2}+2}{(x-2)^{2}(x+1)} = \frac{A}{x-2} + \frac{B}{(x-2)^{2}} + \frac{C}{x+1}\)

Note the crucial difference: we need both \(\frac{A}{x-2}\) AND \(\frac{B}{(x-2)^{2}}\). The term \(\frac{B}{(x-2)^{2}}\) alone is not sufficient.

注意这一关键区别：我们既需要 \(\frac{A}{x-2}\)，也需要 \(\frac{B}{(x-2)^{2}}\)。仅有 \(\frac{B}{(x-2)^{2}}\) 是不够的。

Multiplying through by (x−2)²(x+1) gives:

两边乘以 (x−2)²(x+1) 得：

\(3x^{2}+2 = A(x−2)(x+1) + B(x+1) + C(x−2)^{2}\)

Substitute x = 2: 3(4) + 2 = B(3), giving B = 14⁄3.

代入 x = 2：3(4) + 2 = B(3)，得 B = 14⁄3。

Substitute x = −1: 3(1) + 2 = C(−3)², so C = 5⁄9.

代入 x = −1：3(1) + 2 = C(−3)²，所以 C = 5⁄9。

For A, compare coefficients of x² (or substitute any other value, say x = 0). Equating x² coefficients: 3 = A + C, so A = 3 − 5⁄9 = 22⁄9.

求 A 时，比较 x² 的系数（或代入任意其他值，如 x = 0）。令 x² 系数相等：3 = A + C，所以 A = 3 − 5⁄9 = 22⁄9。

\(\frac{3x^{2}+2}{(x-2)^{2}(x+1)} = \frac{22}{9(x-2)} + \frac{14}{3(x-2)^{2}} + \frac{5}{9(x+1)}\)

In general, for a factor (ax + b)ⁿ, include terms \(\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^{2}} + \dots + \frac{A_n}{(ax+b)^{n}}\).

一般来说，对于因子 (ax + b)ⁿ，需包含项 \(\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^{2}} + \dots + \frac{A_n}{(ax+b)^{n}}\)。

4. Case 3: Quadratic Factors | 情形三：二次因子

If the denominator contains an irreducible quadratic factor (one that cannot be factored into real linear factors), such as x² + 1 or x² + x + 1, the partial fraction includes a linear numerator.

如果分母含有不可约二次因子（即无法分解为实线性因子的二次多项式），如 x² + 1 或 x² + x + 1，则部分分数包含一次分子。

For the fraction \(\frac{2x+3}{(x-1)(x^{2}+1)}\), we write:

对于分数 \(\frac{2x+3}{(x-1)(x^{2}+1)}\)，我们写作：

\(\frac{2x+3}{(x-1)(x^{2}+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^{2}+1}\)

The numerator Bx + C is linear because the denominator is quadratic. Multiplying through by (x−1)(x²+1):

分子 Bx + C 是线性的，因为分母是二次的。两边乘以 (x−1)(x²+1)：

\(2x+3 = A(x^{2}+1) + (Bx+C)(x-1)\)

Substitute x = 1: 2(1) + 3 = A(2), so A = 5⁄2.

代入 x = 1：2(1) + 3 = A(2)，所以 A = 5⁄2。

To find B and C, compare coefficients. Expanding the right-hand side: Ax² + A + Bx² − Bx + Cx − C = (A+B)x² + (C−B)x + (A−C).

为了求 B 和 C，比较系数。展开右边：Ax² + A + Bx² − Bx + Cx − C = (A+B)x² + (C−B)x + (A−C)。

Equating coefficients with 0x² + 2x + 3:

与 0x² + 2x + 3 比较系数：

x²: A + B = 0, so B = −5⁄2
x: C − B = 2, so C = 2 + (−5⁄2) = −1⁄2
constant: A − C = 3 ✓

x²：A + B = 0，所以 B = −5⁄2
x：C − B = 2，所以 C = 2 + (−5⁄2) = −1⁄2
常数项：A − C = 3 ✓

\(\frac{2x+3}{(x-1)(x^{2}+1)} = \frac{5}{2(x-1)} – \frac{5x+1}{2(x^{2}+1)}\)

For a repeated quadratic factor like (x²+1)², include both \(\frac{Bx+C}{x^{2}+1}\) and \(\frac{Dx+E}{(x^{2}+1)^{2}}\).

对于重复二次因子如 (x²+1)²，需同时包含 \(\frac{Bx+C}{x^{2}+1}\) 和 \(\frac{Dx+E}{(x^{2}+1)^{2}}\)。

5. Improper Fractions: Long Division First | 假分数：先做长除法

A rational expression is proper if the degree of the numerator is less than the degree of the denominator. If the degree of the numerator is greater than or equal to that of the denominator, it is improper, and we must perform polynomial long division before decomposing into partial fractions.

如果分子的次数小于分母的次数，则该有理表达式是真分数；如果分子的次数大于或等于分母的次数，则为假分数，必须先进行多项式长除法，再进行部分分数分解。

Consider \(\frac{x^{3}+2x^{2}+1}{x^{2}-1}\). The numerator has degree 3, which is greater than the denominator’s degree 2.

考虑 \(\frac{x^{3}+2x^{2}+1}{x^{2}-1}\)。分子次数为3，大于分母次数2。

Performing long division:

进行长除法：

\(x^{3}+2x^{2}+1 = (x+2)(x^{2}-1) + (x+3)\)

\(\frac{x^{3}+2x^{2}+1}{x^{2}-1} = x + 2 + \frac{x+3}{x^{2}-1}\)

Now decompose the proper remainder \(\frac{x+3}{x^{2}-1}\):

现在对余下的真分数 \(\frac{x+3}{x^{2}-1}\) 进行分解：

\(\frac{x+3}{x^{2}-1} = \frac{x+3}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}\)

Solving gives A = 2 and B = −1. Thus the full answer is:

解得 A = 2，B = −1。因此完整答案为：

\(x + 2 + \frac{2}{x-1} – \frac{1}{x+1}\)

Always check the degrees first — this step is frequently overlooked and causes many exam errors.

务必先检查次数——这一步经常被忽略，导致许多考试错误。

6. The Cover-Up Rule | 覆盖法（遮挡法）

The cover-up rule is a shortcut for finding constants when dealing with distinct linear factors. To find the constant associated with the factor (x − a), simply “cover up” (x − a) in the original expression and evaluate the remaining fraction at x = a.

覆盖法是处理不同线性因子时求常数的捷径。要求因子 (x − a) 对应的常数，只需在原表达式中“遮住”(x − a)，然后在 x = a 处计算剩余分数的值。

Take \(\frac{3x+5}{(x-2)(x+4)}\). To find the constant for \(\frac{A}{x-2}\), cover (x−2) and substitute x = 2:

以 \(\frac{3x+5}{(x-2)(x+4)}\) 为例。要求 \(\frac{A}{x-2}\) 的常数 A，遮住 (x−2) 并代入 x = 2：

\(A = \frac{3(2)+5}{2+4} = \frac{11}{6}\)

Similarly, for B, cover (x+4) and substitute x = −4:

类似地，求 B 时遮住 (x+4) 并代入 x = −4：

\(B = \frac{3(-4)+5}{-4-2} = \frac{-7}{-6} = \frac{7}{6}\)

The cover-up rule does not work directly with repeated factors or irreducible quadratics, but it is an excellent time-saver for the simplest cases.

覆盖法不适用于重复因子或不可约二次因子，但它在最简单的情形下是极佳的时间节省技巧。

7. Comparing Coefficients vs Substitution | 系数比较法与代入法

Two main methods can be used to solve for unknown constants: substitution of x-values and comparison of coefficients. Each has its advantages depending on the situation.

求解未知常数有两种主要方法：代入 x 值和比较系数。两者各有优势，取决于具体情形。

Method \| 方法	When to Use \| 适用场景	Advantage \| 优势
Substitution 代入法	Distinct linear factors 不同线性因子	Fast; each constant isolated 快速，每个常数独立求解
Coefficients 系数法	Repeated factors / quadratics 重复因子 / 二次因子	Systematic, works universally 系统性强，通用性好

In practice, a hybrid approach works best: substitute the roots of the denominator to eliminate terms, then compare one or two remaining coefficients. This combines the speed of substitution with the completeness of coefficient comparison.

在实践中，混合方法效果最佳：先代入分母的根以消去某些项，再比较剩余的一两个系数。这结合了代入法的速度与系数比较法的完备性。

8. Applications: Integration | 应用：积分

One of the most common reasons for using partial fractions in A-Level is integration. Rational functions are often impossible to integrate in their original form, but become straightforward once decomposed.

在A-Level中使用部分分数最常见的理由之一是积分。有理函数通常无法以原形式直接积分，但分解后就变得很简单。

Consider \(\int \frac{5x-2}{(x-1)(x-3)}\,dx\). From Section 2, we know:

考虑 \(\int \frac{5x-2}{(x-1)(x-3)}\,dx\)。由第2节可知：

\(\int \left( \frac{-3}{2(x-1)} + \frac{13}{2(x-3)} \right) dx = -\frac{3}{2}\ln|x-1| + \frac{13}{2}\ln|x-3| + C\)

The result follows directly from the standard rule \(\int \frac{k}{ax+b}\,dx = \frac{k}{a}\ln|ax+b| + C\).

该结果直接来自标准法则 \(\int \frac{k}{ax+b}\,dx = \frac{k}{a}\ln|ax+b| + C\)。

For repeated factors, use the rule \(\int (ax+b)^{-n}\,dx = \frac{(ax+b)^{-n+1}}{a(-n+1)} + C\) (for n ≠ 1).

对于重复因子，使用法则 \(\int (ax+b)^{-n}\,dx = \frac{(ax+b)^{-n+1}}{a(-n+1)} + C\)（n ≠ 1 时）。

9. Applications: Binomial Expansion | 应用：二项式展开

Partial fractions also enable binomial expansions of rational functions. When a fraction is decomposed into simpler pieces, each piece can be expanded using the binomial series, then combined to give a valid series expansion.

部分分数还能实现有理函数的二项式展开。当一个分数被分解为更简单的部分后，每一部分都可以用二项式级数展开，然后合并得到有效的级数展开。

Suppose we wish to expand \(\frac{1}{(1-x)(1+2x)}\) in ascending powers of x. First decompose:

假设我们要将 \(\frac{1}{(1-x)(1+2x)}\) 按 x 的升幂展开。首先分解：

\(\frac{1}{(1-x)(1+2x)} = \frac{A}{1-x} + \frac{B}{1+2x}\)

Solving gives A = 1⁄3, B = 2⁄3. Then expand each term:

解得 A = 1⁄3，B = 2⁄3。然后展开每一项：

\(\frac{1}{3}(1-x)^{-1} + \frac{2}{3}(1+2x)^{-1}\)

Using \((1+u)^{-1} = 1 – u + u^2 – u^3 + \dots\):

利用 \((1+u)^{-1} = 1 – u + u^2 – u^3 + \dots\)：

\(\frac{1}{3}(1 + x + x^2 + x^3 + \dots) + \frac{2}{3}(1 – 2x + 4x^2 – 8x^3 + \dots)\)

\(= 1 – x + 3x^2 – 5x^3 + \dots\)

This technique is essential for finding series expansions that are used in numerical approximation and modelling.

该技巧对于求级数展开至关重要，广泛应用于数值逼近和建模。

10. Common Mistakes and Pitfalls | 常见错误与陷阱

Students frequently make several avoidable errors when dealing with partial fractions. Being aware of these can significantly improve accuracy in exams.

学生在处理部分分数时经常犯几个可以避免的错误。了解这些错误能显著提高考试中的准确率。

Forgetting to check if the fraction is proper before decomposing.
Omitting the constant term Bx + C for quadratic factors — a linear numerator is always required for a quadratic denominator.
Not including all powers for repeated factors, e.g., writing only \(\frac{A}{(x-2)^{2}}\) without \(\frac{B}{x-2}\).
Algebraic sign errors in the substitution step, especially with negative x-values.

忘记在分解前检查分数是否为真分数。
对于二次因子遗漏常数项之外的 Bx + C 项——二次分母必须配一次分子。
对于重复因子未包含所有幂次，例如只写 \(\frac{A}{(x-2)^{2}}\) 而缺少 \(\frac{B}{x-2}\)。
代入步骤中的符号错误，尤其是当 x 取负值时。

A systematic approach — check degree, factor denominator, write correct form, solve carefully — eliminates most of these issues.

系统化的流程——检查次数、分解分母、写出正确形式、仔细求解——能消除大多数这些问题。

11. Exam Tips and Worked Strategy | 考试技巧与解题策略

In examinations, partial fraction questions often appear within larger problems on integration or binomial expansion. A clear strategy is essential.

在考试中，部分分数问题通常出现在积分或二项式展开的综合性题目中。清晰的策略至关重要。

Follow this five-step framework:

遵循以下五步框架：

Check degrees: If the numerator’s degree ≥ denominator’s degree, perform long division first.
Factor the denominator: Fully factorise into linear and irreducible quadratic factors.
Write the correct form: Set up the decomposition template with unknown constants.
Solve for constants: Use substitution at roots, then compare coefficients for remaining unknowns.
Verify: Substitute a simple x-value (like x = 0) to check the identity holds.

检查次数：若分子次数 ≥ 分母次数，先做长除法。
分解分母：将分母完全分解为线性因子和不可约二次因子。
写出正确形式：设定含未知常数的分解模板。
求常数：先在根处代入，再比较剩余系数。
验证：代入一个简单 x 值（如 x = 0）检验恒等式是否成立。

Time management tip: partial fraction decomposition typically carries 4–6 marks. Allocate no more than 5–8 minutes to this stage, leaving ample time for the subsequent integration or expansion.

时间管理建议：部分分数分解通常占 4–6 分。此阶段建议控制在 5–8 分钟内，为后续积分或展开留出充分时间。

12. Conclusion | 总结

Partial fractions is a cornerstone technique in A-Level Mathematics. Mastering the three main cases — distinct linear factors, repeated linear factors, and quadratic factors — along with proper handling of improper fractions, unlocks a wide range of problems in calculus and series work.

部分分数是A-Level数学中的基石技巧。掌握三大主要情形——不同线性因子、重复线性因子和二次因子——以及正确处理假分数，将为微积分和级数领域的广泛问题扫清障碍。

Practice is the key. Work through problems from past papers, always setting up the correct form before solving for constants. With consistent effort, partial fractions will become a reliable and even enjoyable part of your mathematical toolkit.

练习是关键。认真研究历年真题，始终先写出正确的分解形式再求解常数。只要持之以恒，部分分数将成为你数学工具箱中可靠甚至令人愉悦的工具。

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Partial Fractions: Splitting and Simplifying | 部分分数的拆分与化简