Partial Fractions | 部分分式分解

📚 Partial Fractions | 部分分式分解

Partial fractions is a fundamental algebraic technique used to express a complex rational function as a sum of simpler fractions. This decomposition is extremely useful in integration, differentiation, series expansion, and solving differential equations. In the IB Mathematics curriculum, mastering partial fractions is essential for both Analysis & Approaches (AA) and Applications & Interpretation (AI) courses, especially when dealing with rational expressions that cannot be simplified directly. This article will guide you through the entire process, from identifying proper and improper fractions to working with distinct linear factors, repeated factors, and irreducible quadratics.

部分分式是一种基本的代数技巧，用于将复杂的有理函数表示为几个简单分式的和。这种分解在积分、微分、级数展开以及解微分方程中极为有用。在IB数学课程中，无论是分析与方法(AA)还是应用与解释(AI)，掌握部分分式都至关重要，特别是在处理无法直接化简的有理表达式时。本文将带你走完整个过程，从识别真分式与假分式，到处理不同线性因式、重复因式和不可约二次因式。

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

A rational function is a ratio of two polynomials, such as (2x+3)/(x² – 1). Partial fraction decomposition rewrites this single fraction into a sum of two or more simpler fractions with denominators corresponding to the factors of the original denominator. The idea is to reverse the process of adding algebraic fractions. For example, 3/(x–1) + 1/(x+2) can be combined into a single fraction; partial fractions allow us to go backwards from that single fraction to the sum of its components.

有理函数是两个多项式的比，例如 (2x+3)/(x² – 1)。部分分式分解将这个单一分式改写为两个或多个更简单分式的和，其分母与原分母的因式一一对应。其思想是代数分式加法的逆过程。例如，3/(x–1) + 1/(x+2) 可以合并为一个分式；部分分式则让我们从一个合成好的分式返回到各组成部分之和。

This technique heavily relies on the fact that any rational function with a factorable denominator can be uniquely decomposed, provided the original fraction is in its proper form — meaning the degree of the numerator is less than the degree of the denominator.

该技巧高度依赖于这样一个事实：任何分母可分解因式的有理函数，只要原本是真分式（即分子的次数低于分母的次数），就可以被唯一地分解。

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2. Proper and Improper Rational Functions | 真分式与假分式

Before applying partial fractions, we must check whether the rational expression is proper. A proper fraction has the degree of the numerator strictly less than the degree of the denominator. For instance, (x+1)/(x²+3x+2) is proper because numerator degree is 1 and denominator degree is 2. If the numerator's degree is equal to or greater than the denominator's, the fraction is improper. An example is (x²+2)/(x²–1), where both degrees are 2.

在应用部分分式之前，我们必须检查有理表达式是否为真分式。真分式要求分子的次数严格小于分母的次数。例如，(x+1)/(x²+3x+2) 是真分式，因为分子次数为1，分母次数为2。如果分子的次数大于或等于分母的次数，该分式就是假分式。例如 (x²+2)/(x²–1)，分子和分母的次数都是2。

When faced with an improper fraction, we must first perform polynomial long division to express it as a polynomial plus a proper rational fraction. Only the proper part is then decomposed into partial fractions. This step is crucial because applying the decomposition directly to an improper fraction leads to an incorrect or incomplete result.

遇到假分式时，必须先进行多项式长除法，将其表示为一个多项式与一个真分式之和。然后只对真分式部分进行部分分式分解。这一步至关重要，因为直接对假分式分解会得到错误或不完整的结果。

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3. Distinct Linear Factors | 不同线性因式

The simplest case occurs when the denominator factorises into distinct linear factors, i.e., factors of the form (ax+b) with no repetition. Suppose the denominator is (a₁x+b₁)(a₂x+b₂)…(aₙx+bₙ). Then we set up the decomposition as a sum of fractions, each with a constant numerator: A₁/(a₁x+b₁) + A₂/(a₂x+b₂) + … + Aₙ/(aₙx+bₙ).

最简单的情况出现在分母可分解为不同的线性因式时，即所有因式都是 (ax+b) 的形式且没有重复。假设分母为 (a₁x+b₁)(a₂x+b₂)…(aₙx+bₙ)，那么我们将分解设为一系列分式之和，每个分子都是常数：A₁/(a₁x+b₁) + A₂/(a₂x+b₂) + … + Aₙ/(aₙx+bₙ)。

The constants A₁, A₂, …, Aₙ are found by multiplying both sides by the original denominator, equating coefficients of like powers of x, or substituting convenient values of x that make some factors zero. The substitution method is often faster. After finding the constants, we can rewrite the original expression in its decomposed form.

常数 A₁、A₂、…、Aₙ 可以通过将等式两边同乘原分母，然后对比同次幂系数，或者代入使某些因式为零的方便 x 值来求得。代入法通常更快。求出常数后，即可将原表达式写成分解形式。

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4. Example: Distinct Linear Factors | 示例：不同线性因式

Decompose the rational expression:

(5x+1)/((x–1)(x+2))

分解以下有理表达式：

(5x+1)/((x–1)(x+2))

Since the denominator has two distinct linear factors, we write:

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

因为分母有两个不同的线性因式，我们设：

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

Multiply both sides by (x–1)(x+2) to clear denominators:

5x + 1 = A(x+2) + B(x–1)

两边同乘 (x–1)(x+2) 去掉分母：

5x + 1 = A(x+2) + B(x–1)

Choose values that simplify the equation: let x = 1, then the term with B vanishes:

5(1)+1 = A(1+2) -> 6 = 3A => A = 2

选择能简化方程的值：令 x = 1，含 B 的项消失：

5(1)+1 = A(1+2) → 6 = 3A ⇒ A = 2

Let x = –2, the term with A vanishes:

5(–2)+1 = B(–2–1) -> –9 = –3B => B = 3

令 x = –2，含 A 的项消失：

5(–2)+1 = B(–2–1) → –9 = –3B ⇒ B = 3

Thus the decomposition is:

(5x+1)/((x–1)(x+2)) = 2/(x–1) + 3/(x+2)

因此分解结果为：

(5x+1)/((x–1)(x+2)) = 2/(x–1) + 3/(x+2)

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5. Repeated Linear Factors | 重复线性因式

When a linear factor is repeated, e.g., (ax+b)ⁿ, the decomposition must include a series of terms for that factor with increasing powers in the denominator. For a factor repeated n times, we use A₁/(ax+b) + A₂/(ax+b)² + … + Aₙ/(ax+b)ⁿ. This ensures we have enough constants to cover all possible remainders in the original function.

当某个线性因式重复出现时，例如 (ax+b)ⁿ，分解中必须包含针对该因式的一系列项，分母次数递增。对于一个重复 n 次的因式，我们使用 A₁/(ax+b) + A₂/(ax+b)² + … + Aₙ/(ax+b)ⁿ。这样保证有足够的常数覆盖原函数中所有可能的余项。

It is essential to include every power from 1 up to n; skipping a power would make it impossible to match the original numerator. The constants can again be determined by multiplying through by the full denominator and comparing coefficients or substituting values, though we often combine both methods to handle multiple constants efficiently.

必须包含从1次幂到n次幂的每一项；漏掉某一幂次将无法与原始分子匹配。常数同样可通过乘以整个分母并比较系数或代入值来确定，不过我们通常会结合两种方法以高效地处理多个常数。

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6. Example: Repeated Linear Factors | 示例：重复线性因式

Decompose:

(2x+1)/(x–1)²

分解：

(2x+1)/(x–1)²

Here (x–1) is repeated twice, so we set up:

(2x+1)/(x–1)² = A/(x–1) + B/(x–1)²

这里 (x–1) 重复两次，因此设：

(2x+1)/(x–1)² = A/(x–1) + B/(x–1)²

Multiply both sides by (x–1)²:

2x + 1 = A(x–1) + B

两边同乘 (x–1)²：

2x + 1 = A(x–1) + B

Expanding: 2x+1 = Ax – A + B. Comparing coefficients of x gives A = 2. Comparing the constant term: 1 = –A + B, so B = 1 + A = 3.

展开：2x+1 = Ax – A + B。比较 x 的系数得 A = 2。比较常数项：1 = –A + B，故 B = 1 + A = 3。

ields B: 2(1)+1 = A(0) + B => B = 3. Then substituting any other value, say x=0, gives 1 = –A + 3 => A = 2. The decomposition is:

(2x+1)/(x–1)² = 2/(x–1) + 3/(x–1)²

另一种方法，代入 x=1 直接得到 B：2(1)+1 = A(0)+B ⇒ B=3。再代入另一个值，如 x=0，得 1 = –A + 3 ⇒ A=2。分解结果为：

(2x+1)/(x–1)² = 2/(x–1) + 3/(x–1)²

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7. Irreducible Quadratic Factors | 不可约二次因式

When the denominator contains an irreducible quadratic factor ax²+bx+c (i.e., its discriminant b²–4ac < 0 and it cannot be factored into real linear factors), the corresponding numerator in the decomposition is a linear expression, not just a constant. Hence for each distinct irreducible quadratic factor, we include a term (Ax+B)/(ax²+bx+c).

当分母包含不可约二次因式 ax²+bx+c（即判别式 b²–4ac 小于零，无法在实数范围分解为线性因式）时，分解中对应的分子应是一个一次式，而非单纯的常数。因此，对每个不同的不可约二次因式，我们添加一项 (Ax+B)/(ax²+bx+c)。

If the irreducible quadratic factor is repeated, we similarly use a chain of terms with increasing powers, each having a linear numerator. This parallels the repeated linear case but with two unknown coefficients per term. The determination of constants often relies more on equating coefficients because substituting specific x-values does not eliminate the quadratic factor as cleanly.

如果不可约二次因式重复出现，同样要使用一串递增幂次的项，每项都有一个一次分子。这与重复线性因式的情形类似，但每项有两个未知系数。常数的确定更多依赖比较系数法，因为代入具体的 x 值不能干净地消去二次因式。

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8. Example: Quadratic Factors | 示例：二次因式

Decompose:

(x+2)/((x²+1)(x–1))

分解：

(x+2)/((x²+1)(x–1))

The factor x²+1 is irreducible (discriminant negative). Set up:

(x+2)/((x²+1)(x–1)) = (Ax+B)/(x²+1) + C/(x–1)

因式 x²+1 是不可约的（判别式负）。设：

(x+2)/((x²+1)(x–1)) = (Ax+B)/(x²+1) + C/(x–1)

Multiply by the common denominator:

x + 2 = (Ax+B)(x–1) + C(x²+1)

乘以公分母：

x + 2 = (Ax+B)(x–1) + C(x²+1)

C: 1+2 = C(1+1) => 3 = 2C => C = 3/2. Now expand and collect coefficients. Expand (Ax+B)(x–1) = Ax² – Ax + Bx – B. Then add C(x²+1) = (3/2)x² + 3/2. Sum: x² terms: A + 3/2; x terms: –A + B; constants: –B + 3/2. Equate to 0x² + 1x + 2.

令 x = 1 求 C：1+2 = C(1+1) ⇒ 3 = 2C ⇒ C = 3/2。现在展开并收集系数。展开 (Ax+B)(x–1) = Ax² – Ax + Bx – B。加上 C(x²+1) = (3/2)x² + 3/2。总和：x² 项：A + 3/2；x 项：–A + B；常数：–B + 3/2。与 0x² + 1x + 2 对应相等。

From x²: A + 3/2 = 0 ⇒ A = –3/2. From x: –A + B = 1 ⇒ –(–3/2) + B = 1 ⇒ 3/2 + B = 1 ⇒ B = –1/2. Check constant: –B + 3/2 = 1/2 + 3/2 = 2, correct. Finally,

(x+2)/((x²+1)(x–1)) = (–3/2 x – 1/2)/(x²+1) + (3/2)/(x–1)

由 x²: A + 3/2 = 0 ⇒ A = –3/2。由 x: –A + B = 1 ⇒ 3/2 + B = 1 ⇒ B = –1/2。检验常数：–B + 3/2 = 1/2 + 3/2 = 2，正确。最终：

(x+2)/((x²+1)(x–1)) = (–3/2 x – 1/2)/(x²+1) + (3/2)/(x–1)

This can also be written as –(3x+1)/(2(x²+1)) + 3/(2(x–1)).

这也可以写成 –(3x+1)/(2(x²+1)) + 3/(2(x–1))。

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9. Improper Fractions and Polynomial Long Division | 假分式与多项式长除法

inator, with R(x) having degree less than the denominator. Only the proper part R(x)/denominator is then decomposed.

当分子的次数大于或等于分母的次数时，不能直接分解。我们必须进行多项式长除法，得到商多项式 Q(x) 和余式 R(x)，使得原式 = Q(x) + R(x)/分母，且 R(x) 的次数低于分母。然后只对真分式部分 R(x)/分母进行分解。

For example, (x³+x)/(x²–1) is improper. Division yields: x³+x divided by x²–1 gives quotient x and remainder 2x. So (x³+x)/(x²–1) = x + (2x)/(x²–1). Now decompose (2x)/((x–1)(x+1)) into partial fractions. This gives x + 1/(x–1) + 1/(x+1).

例如，(x³+x)/(x²–1) 是假分式。长除法：x³+x 除以 x²–1 得商 x，余 2x。因此 (x³+x)/(x²–1) = x + (2x)/(x²–1)。再对 (2x)/((x–1)(x+1)) 进行部分分式分解，得 x + 1/(x–1) + 1/(x+1)。

Always perform the division first; otherwise the partial fraction set-up will be incorrect because it assumes a proper fraction. This is a common pitfall in exams.

务必先做除法；否则部分分式的设定会因假设为真分式而错误。这是考试中常见的陷阱。

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10. Applications to Integration | 积分应用

typically becomes a simple logarithmic or arctangent integral. For instance, ∫ (5x+1)/((x–1)(x+2)) dx becomes ∫ 2/(x–1) dx + ∫ 3/(x+2) dx = 2 ln|x–1| + 3 ln|x+2| + C.

IB数学中部分分式的主要应用之一是对有理函数积分。分解后，每一项通常变为简单的对数或反正切积分。例如，∫ (5x+1)/((x–1)(x+2)) dx 变为 ∫ 2/(x–1) dx + ∫ 3/(x+2) dx = 2 ln|x–1| + 3 ln|x+2| + C。

For irreducible quadratic denominators, integrals often involve completing the square and lead to arctan formulas. Decomposing ahead of integration turns a complex rational function into a sum of basic forms that are directly integrable, saving time and reducing mistakes.

对于不可约二次分母，积分常通过配方转化为反正切公式。先分解再积分，能将复杂的有理函数化为可直接积分的几个基本形式之和，既节省时间又减少错误。

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11. Applications to Binomial Expansion | 二项展开应用

Partial fractions also facilitate the expansion of rational functions into infinite series. By decomposing a rational expression, each term can often be expressed in the form (ax+b)⁻ⁿ and then expanded using the binomial theorem for negative or fractional powers. This is particularly useful when finding power series representations.

部分分式还有助于将有理函数展开为无穷级数。通过分解有理表达式，每一项常可表示为 (ax+b)⁻ⁿ 的形式，然后用二项式定理对负指数或分数指数进行展开。这在求幂级数表示时特别有用。

For example, after decomposing 2/(x–1) + 3/(x+2) we can write each term as a series expansion valid for certain ranges of x. This method appears frequently in IB questions linking algebra and calculus.

例如，将 2/(x–1) + 3/(x+2) 分解后，我们可以把每一项写作在某 x 范围内成立的级数展开。这种方法经常出现在将代数与微积分联系起来的IB考题中。

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12. Common Mistakes and Tips | 常见错误与提示

Many students forget to check if the fraction is proper before starting the decomposition, leading to an incomplete set-up. Always compare the degrees of numerator and denominator. Another typical mistake is using the wrong form for the numerator when an irreducible quadratic factor is present; remember it must be Ax+B, not a single constant.

许多学生忘记在开始分解前检查分式是否为真分式，导致设定不完整。务必比较分子与分母的次数。另一个典型错误是当存在不可约二次因式时，分子形式用错；记住必须用 Ax+B，而非一个单一的常数。

When solving for constants, students sometimes rely solely on substitution without verifying with coefficient comparison, which can hide arithmetic errors. Use a mixed approach for validation. Also, be careful with signs when expanding and equating. Practice a variety of examples to build confidence, including improper fractions and mixed factor types.

在求解常数时，有些学生只依赖代入法而不通过比较系数验证，可能隐藏计算错误。建议结合两种方法相互校验。此外，在展开和等式移项时要仔细处理符号。通过练习包含假分式与混合因式的各类例题来建立信心。

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