📚 Partial Fraction Decomposition | IB数学：部分分式分解方法
Partial fraction decomposition is one of the most powerful algebraic techniques in the IB Mathematics curriculum. It transforms a complicated rational expression into a sum of simpler fractions, making integration, differentiation, and equation solving significantly more manageable. This skill appears consistently in both Analysis and Approaches (AA) and Applications and Interpretation (AI) courses.
部分分式分解是IB数学课程中最强大的代数技巧之一。它将复杂的有理表达式转化为若干更简单分式的和，使积分、求导和解方程变得容易得多。这项技能在分析与方法（AA）和应用与解释（AI）两门课程中都会稳定出现。
1. What is a Rational Function? | 什么是有理函数？
A rational function is defined as the ratio of two polynomials, written as \( f(x) = \frac{P(x)}{Q(x)} \), where both P(x) and Q(x) are polynomials and Q(x) ≠ 0. In partial fraction decomposition, we require that the degree of P(x) is strictly less than the degree of Q(x). If this is not the case, we must first perform polynomial long division to reduce the fraction to proper form.
有理函数定义为两个多项式的比值，写作 f(x) = P(x) / Q(x)，其中 P(x) 和 Q(x) 都是多项式且 Q(x) ≠ 0。在进行部分分式分解时，我们要求 P(x) 的次数严格小于 Q(x) 的次数。如果情况不是这样，我们必须先用多项式长除法将分式化为真分式形式。
Example | 示例: \(\frac{x^3+1}{x^2-1}\) → 先进行除法 → \(x + \frac{x+1}{x^2-1}\)
2. Why Decompose at All? | 为什么要分解？
Rational functions in their original form are often difficult to integrate. For instance, \(\int \frac{1}{x^2-1}\,dx\) appears challenging at first glance. However, once we decompose: \(\frac{1}{x^2-1} = \frac{1}{2(x-1)} – \frac{1}{2(x+1)}\), the integral becomes two simple natural logarithm terms. Similarly, in differential equations and recurrence relations, partial fractions enable us to isolate variable terms and solve more efficiently.
原始形式的有理函数往往难以积分。例如，∫ 1/(x²−1) dx 乍看之下很有挑战性。然而，一旦我们分解：1/(x²−1) = 1/[2(x−1)] − 1/[2(x+1)]，积分就变成了两个简单的自然对数项。类似地，在微分方程和递推关系中，部分分式使我们能够分离变量项并更高效地求解。
3. Case 1: Distinct Linear Factors | 情形1：不同线性因子
When the denominator Q(x) factors into distinct linear factors, each factor contributes one term to the decomposition. Suppose Q(x) = (x − a₁)(x − a₂)…(x − aₙ), where all aᵢ are distinct. Then:
当分母 Q(x) 分解为互不相同的线性因子时，每个因子在分解式中贡献一项。设 Q(x) = (x − a₁)(x − a₂)…(x − aₙ)，其中所有 aᵢ 互不相同，则：
\(\frac{P(x)}{Q(x)} = \frac{A₁}{x-a₁} + \frac{A₂}{x-a₂} + \cdots + \frac{Aₙ}{x-aₙ}\)
To determine the constants Aᵢ, multiply both sides by Q(x) and then equate coefficients or substitute convenient x-values. The substitution method is particularly elegant: set x = aᵢ, and all terms except the one with Aᵢ vanish.
为确定常数 Aᵢ，将等式两边乘以 Q(x)，然后比较系数或代入方便的x值。代入法尤其巧妙：令 x = aᵢ，则除含 Aᵢ 的项之外的所有项都消去。
Worked Example | 例题： Decompose \(\frac{3x+5}{(x-1)(x+2)}\). Set \(\frac{3x+5}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}\). Multiply through: 3x + 5 = A(x + 2) + B(x − 1). Substituting x = 1 gives 8 = 3A → A = 8/3. Substituting x = −2 gives −1 = −3B → B = 1/3.
例题： 分解 3x+5 除以 (x−1)(x+2)。设该表达式 = A/(x−1) + B/(x+2)。两边同乘得：3x + 5 = A(x + 2) + B(x − 1)。代入 x = 1 得 8 = 3A → A = 8/3。代入 x = −2 得 −1 = −3B → B = 1/3。
4. The Cover-Up Method | 覆盖法
The cover-up rule is a rapid shortcut for Case 1. To find Aᵢ in \(\frac{P(x)}{(x-a₁)(x-a₂)\cdots}\), simply cover the factor (x − aᵢ) in the denominator and evaluate the remaining expression at x = aᵢ. This technique works because multiplying by (x − aᵢ) and then setting x = aᵢ isolates exactly the constant Aᵢ.
覆盖法是针对情形1的快速技巧。要求 Aᵢ 在 P(x)/[(x−a₁)(x−a₂)…] 中的值，只需在分母中盖住因子 (x − aᵢ)，再将剩余表达式在 x = aᵢ 处求值。该技巧之所以有效，是因为乘以 (x − aᵢ) 后令 x = aᵢ 恰好能分离出常数 Aᵢ。
For \(\frac{2x-1}{(x-3)(x+4)}\): cover (x−3) → evaluate at x=3 → \(A = \frac{2(3)-1}{3+4} = \frac{5}{7}\)
5. Case 2: Repeated Linear Factors | 情形2：重复线性因子
When a linear factor (x − a) appears with multiplicity k, the decomposition must include k terms: \(\frac{A₁}{x-a} + \frac{A₂}{(x-a)²} + \cdots + \frac{Aₖ}{(x-a)ᵏ}\). Students frequently make the mistake of writing only a single term—this is the most common error in this topic.
当线性因子 (x − a) 以重数 k 出现时，分解式中必须包含 k 项：A₁/(x−a) + A₂/(x−a)² + … + Aₖ/(x−a)ᵏ。学生常犯的错误是只写一项——这是本主题中最常见的错误。
Worked Example | 例题： Decompose \(\frac{x²+2x+3}{(x-1)³}\). Write: \(\frac{A}{x-1} + \frac{B}{(x-1)²} + \frac{C}{(x-1)³}\). Multiply through by (x−1)³: x² + 2x + 3 = A(x−1)² + B(x−1) + C. Expand: A(x²−2x+1) + B(x−1) + C. Equating coefficients: A = 1, −2A + B = 2 → B = 4, A − B + C = 3 → C = 6.
例题： 分解 (x²+2x+3)/(x−1)³。设 = A/(x−1) + B/(x−1)² + C/(x−1)³。两边同乘 (x−1)³：x² + 2x + 3 = A(x−1)² + B(x−1) + C。展开得 A(x²−2x+1) + B(x−1) + C。比较系数：A = 1，−2A + B = 2 → B = 4，A − B + C = 3 → C = 6。
6. Case 3: Irreducible Quadratic Factors | 情形3：不可约二次因子
If Q(x) contains a quadratic factor ax² + bx + c that has no real roots (i.e., discriminant b² − 4ac < 0), then the corresponding partial fraction term must be of the form \(\frac{Ax + B}{ax²+bx+c}\). Note that the numerator must be linear, not just a constant, because a constant numerator cannot represent all possible rational functions with such a denominator.
若 Q(x) 含有无实根的二次因子 ax² + bx + c（即判别式 b² − 4ac < 0），则对应的部分分式项必须采用 (Ax + B)/(ax² + bx + c) 的形式。注意分子必须是线性的，而不仅仅是常数，因为常数分子无法表示具有这种分母的所有可能有理函数。
Worked Example | 例题： \(\frac{2x+1}{(x²+1)(x-2)} = \frac{Ax+B}{x²+1} + \frac{C}{x-2}\). Multiply through: 2x+1 = (Ax+B)(x−2) + C(x²+1). Expanding: 2x+1 = Ax² − 2Ax + Bx − 2B + Cx² + C. Group: (A+C)x² + (−2A+B)x + (−2B+C). Comparing coefficients: A + C = 0, −2A + B = 2, −2B + C = 1. Solving: from A = −C, substitute to get C = −1, A = 1, B = 4.
例题： (2x+1)/[(x²+1)(x−2)] = (Ax+B)/(x²+1) + C/(x−2)。两边同乘：2x+1 = (Ax+B)(x−2) + C(x²+1)。展开：2x+1 = Ax² − 2Ax + Bx − 2B + Cx² + C。整理：(A+C)x² + (−2A+B)x + (−2B+C)。比较系数：A + C = 0，−2A + B = 2，−2B + C = 1。解得：由 A = −C 代入，得 C = −1，A = 1，B = 4。
7. Repeated Quadratic Factors | 重复的二次因子
When an irreducible quadratic factor appears with multiplicity k, we must include terms of increasing denominator powers: \(\frac{A₁x + B₁}{ax²+bx+c} + \frac{A₂x + B₂}{(ax²+bx+c)²} + \cdots\). Each term’s numerator is linear. This is the most general case and requires careful coefficient comparison, often leading to a system of four or more linear equations.
当不可约二次因子以重数 k 出现时，我们必须包含分母幂次递增的各项：(A₁x+B₁)/(ax²+bx+c) + (A₂x+B₂)/(ax²+bx+c)² + …。每一项的分子都是线性的。这是最一般的情形，需要仔细比较系数，通常会得到包含四个或更多线性方程的方程组。
8. Method of Equating Coefficients | 比较系数法
This is the most systematic approach. After multiplying both sides of the decomposition identity by the full denominator, you expand the right-hand side and group terms by powers of x. You then equate the coefficient of each power on the left with the corresponding coefficient on the right. This produces a system of linear equations in the unknown constants.
这是最系统的方法。将分解恒等式两边乘以完整的分母后，展开右边并按 x 的幂次归组。然后将左边每一项的幂次系数与右边对应系数相等。这样就产生了关于未知常数的线性方程组。
This method is particularly advantageous when dealing with repeated factors or when the substitution method cannot fully determine all constants. It always works, regardless of the complexity of the denominator.
这种方法在处理重复因子或代入法无法完全确定所有常数时特别有优势。无论分母多么复杂，它总是有效的。
9. Improper Fractions and Long Division | 假分式与长除法
When the degree of the numerator is greater than or equal to the degree of the denominator, partial fraction decomposition cannot begin directly. You must first perform polynomial long division to obtain: \(\frac{P(x)}{Q(x)} = \text{quotient} + \frac{\text{remainder}}{Q(x)}\). The quotient is a polynomial, and the remainder term is now a proper fraction that can be decomposed using the methods above.
当分子的次数大于等于分母的次数时，不能直接进行部分分式分解。你必须先进行多项式长除法，得到：P(x)/Q(x) = 商 + 余数/Q(x)。商是一个多项式，余数项现在是一个真分式，可以使用上述方法分解。
Worked Example | 例题： For \(\frac{x³}{x²-1}\), long division gives \(x + \frac{x}{x²-1}\). Then decompose: \(\frac{x}{x²-1} = \frac{1}{2(x-1)} + \frac{1}{2(x+1)}\). Final answer: \(x + \frac{1}{2(x-1)} + \frac{1}{2(x+1)}\).
例题： 对 x³/(x²−1)，长除法得 x + x/(x²−1)。再分解：x/(x²−1) = 1/[2(x−1)] + 1/[2(x+1)]。最终答案：x + 1/[2(x−1)] + 1/[2(x+1)]。
10. Application: Integration | 应用：积分
The most important application in IB is integration. Once a rational function is decomposed, each term can be integrated using standard formulas. For linear denominators: \(\int \frac{A}{x-a}\,dx = A\ln|x-a| + C\). For quadratic denominators: \(\int \frac{Ax+B}{ax²+bx+c}\,dx\) often requires completing the square or recognizing an arctangent form.
在IB中最重要的应用是积分。一旦有理函数被分解，每一项都可以使用标准公式积分。对于线性分母：∫ A/(x−a) dx = A ln|x−a| + C。对于二次分母：∫ (Ax+B)/(ax²+bx+c) dx 通常需要配方法或识别反正切形式。
\(\int \frac{1}{x²-a²}\,dx = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C\)
This formula directly follows from partial fraction decomposition and is explicitly allowed in the IB formula booklet, but understanding its derivation helps with more complex problems.
这个公式直接来源于部分分式分解，在IB公式手册中是明确允许使用的，但理解其推导过程有助于解决更复杂的问题。
11. Application: Differential Equations | 应用：微分方程
In separable differential equations, partial fractions often appear when the equation involves the product of linear expressions. For example, \(\frac{dy}{dx} = y(y-1)\) can be solved by separating variables: \(\int \frac{1}{y(y-1)}\,dy = \int dx\). The left integral requires decomposing \(\frac{1}{y(y-1)} = \frac{-1}{y} + \frac{1}{y-1}\), giving \(-\ln|y| + \ln|y-1| = x + C\).
在可分离变量的微分方程中，当方程涉及线性表达式的乘积时，部分分式经常出现。例如，dy/dx = y(y−1) 可以通过分离变量求解：∫ 1/[y(y−1)] dy = ∫ dx。左边的积分需要分解 1/[y(y−1)] = −1/y + 1/(y−1)，得到 −ln|y| + ln|y−1| = x + C。
12. Common Pitfalls and Exam Tips | 常见误区与考试建议
Students often forget to check whether the rational function is proper before decomposing. Always verify degrees first. Another common mistake is omitting the linear numerator for quadratic factors. Finally, always verify your decomposition by recombining the fractions over a common denominator—this simple check takes only seconds but catches most errors.
学生在分解前常常忘记检查有理函数是否为真分式。务必先验证次数。另一个常见错误是为二次因子遗漏线性分子。最后，始终通过将分式重新合并为公分母来验证你的分解结果——这个简单的检查只需几秒钟，但能发现大多数错误。
- Always check degrees first | 始终先检查次数
- Use the cover-up method for quick checks | 使用覆盖法快速检查
- For quadratic factors, numerator must be linear | 对于二次因子，分子必须是线性的
- Verify by recombining or substituting a test value | 通过重新合并或代入测试值验证
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