📚 14 Partial Fractions | 部分分式
In algebra, a rational function is a ratio of two polynomials. Many problems in calculus, differential equations, and engineering require us to break such a ratio into a sum of simpler fractions called partial fractions. This technique reverses the process of adding algebraic fractions and is essential for integrating rational expressions, solving recurrence relations, and working with Laplace transforms. For IB Mathematics students, mastering partial fractions means being able to decompose a given rational expression into a sum of terms with linear or quadratic denominators, then using those results to integrate, find binomial expansions, or solve equations.
在代数学中，有理函数是两个多项式的比。在微积分、微分方程和工程学的许多问题中，都需要将这样的比式拆分为若干个更简单的分式之和，这些分式就称为部分分式。这一方法相当于把代数分式相加的过程逆过来，对于有理式的积分、递推关系的求解以及拉普拉斯变换等工作来说必不可少。对IB数学学生而言，掌握部分分式意味着能够将给定的有理表达式分解为分母为一次或二次式的多项之和，进而用于积分、二项展开或方程求解。
1. What Are Partial Fractions? | 什么是部分分式？
A partial fraction decomposition expresses a single rational function as a sum of simpler fractions whose denominators are factors of the original denominator. For example, (2x+3)/((x-1)(x+2)) can be written as A/(x-1) + B/(x+2), where A and B are constants to be determined. The method relies on the fact that every proper rational function with a factorisable denominator can be uniquely decomposed, provided we account for the nature of each factor (linear, repeated, or irreducible quadratic).
部分分式分解是指将一个有理函数表示为几个更简单的分式之和，这些简单分式的分母恰为原分母的因式。例如，(2x+3)/((x-1)(x+2)) 可以写成 A/(x-1) + B/(x+2)，其中 A 和 B 是待定常数。该方法依据的基本事实是：每一个分母可因式分解的真分式都可以唯一地拆分成部分分式，前提是要根据每个因式的类型（一次、重复或不可约二次）正确设定分式形式。
The primary goal is to turn a complicated algebraic fraction into a form that is much easier to integrate, differentiate, or expand. In IB examinations, you are typically asked to find the constants A, B, C, ... and then use the decomposition for integration or series expansion.
主要目标是把一个复杂的代数分式变成更易于积分、求导或展开的形式。在IB考试中，通常会要求你求出常数 A、B、C……，然后利用分解结果进行积分或级数展开。
The technique is not limited to pure mathematics; it appears in physics when solving rate equations, in economics when analysing multiplier effects, and in engineering when handling transfer functions.
这一技巧不仅限于纯数学；在物理中解速率方程时、在经济学中分析乘数效应时、在工程学中处理传递函数时都会出现部分分式。
2. Proper and Improper Rational Functions | 真分式与假分式
Before attempting partial fractions, we must ensure the rational expression is proper, meaning the degree of the numerator polynomial is strictly less than the degree of the denominator polynomial. For example, (x+1)/(x²+3x+2) is proper because numerator degree 1 < denominator degree 2. However, (x³+2x)/(x²-1) has numerator degree 3 and denominator degree 2; it is improper.
在尝试部分分式分解前，我们必须确保该有理表达式为真分式，即分子多项式的次数严格小于分母多项式的次数。例如，(x+1)/(x²+3x+2) 是真分式，因为分子次数1 < 分母次数2。而 (x³+2x)/(x²-1) 的分子次数为3，分母次数为2，是一个假分式。
/(x²-1) after long division; the term (3x)/(x²-1) is proper and can be further decomposed.
如果是有理假分式，则需要先进行多项式长除法，得到一个多项式商加上一个真分式余项。然后只对余项进行部分分式分解。例如，通过长除法可将 (x³+2x)/(x²-1) 写成 x + (3x)/(x²-1)；其中 (3x)/(x²-1) 为真分式，可进一步分解。
In IB problems, long division is a common preliminary step. Always check degrees before choosing the partial fraction form. Neglecting this step leads to incorrect equations because the assumed form may not match the original expression's degree.
在IB题目中，长除法是常见的预备步骤。在选择部分分式形式之前，务必检查次数。忽略这一步会导致方程不正确，因为所假设的形式可能不匹配原表达式的次数。
3. Linear Factors (Non-Repeated) | 无重复一次因式
When the denominator factorises into distinct linear factors, the partial fraction form has one term for each factor, with a constant numerator. For a denominator (x-a)(x-b)…, the decomposition is A/(x-a) + B/(x-b) + … . To find A and B, we multiply through by the original denominator and then either substitute convenient x-values or equate coefficients of powers of x.
当分母分解为不同的一次因式时，部分分式的形式为每个因式对应一项，且分子为常数。对于分母 (x-a)(x-b)……，其分解形式为 A/(x-a) + B/(x-b) + ……。为求出 A 和 B，可将等式两边同乘以原分母，然后代入方便的 x 值求解，或者比较 x 的各次幂的系数。
3)-1 = B(-5) → -16 = -5B → B = 16/5. Thus the decomposition is (9/5)/(x-2) + (16/5)/(x+3).
例子：将 (5x-1)/((x-2)(x+3)) 进行分解。设 (5x-1)/((x-2)(x+3)) = A/(x-2) + B/(x+3)。两边同乘 (x-2)(x+3)：5x-1 = A(x+3) + B(x-2)。代入 x=2 以消去 B：5(2)-1 = A(5) → 9 = 5A → A = 9/5。代入 x=-3：5(-3)-1 = B(-5) → -16 = -5B → B = 16/5。因此分解为 (9/5)/(x-2) + (16/5)/(x+3)。
The substitution method is efficient for distinct linear factors. Alternatively, expanding and equating coefficients gives the same results.
对于不同的线性因式，代入法非常高效。另一种方法是展开后比较系数，可以得到相同的结果。
4. Repeated Linear Factors | 重复一次因式
If a linear factor appears raised to a power n, the decomposition must include a chain of fractions with denominators (x-a), (x-a)², …, (x-a)ⁿ. Each term's numerator is a constant. For example, for a denominator containing (x-1)³, the partial fraction form includes A/(x-1) + B/(x-1)² + C/(x-1)³.
如果一个一次因式出现幂次 n，分解时必须包含一串分母为 (x-a)、(x-a)²、……、(x-a)ⁿ 的分式，每一项的分子都是常数。例如，分母中含有 (x-1)³ 时，其部分分式形式应包括 A/(x-1) + B/(x-1)² + C/(x-1)³。
irst to find the highest-power coefficient. Then compare coefficients or substitute other convenient x-values to find the rest.
遗漏较低次幂的项是一个典型错误。原因是代数完备性：这些项之和可以还原 (x-a)ⁿ 之上任意次数不超过 n-1 的多项式分子。为确定常数，先去分母并优先代入 x = a 求出最高次幂的系数，然后再通过比较系数或代入其他方便的 x 值求出其余常数。
2x+1 = Ax - 2A + 5. Equate x coefficients: 2 = A. The decomposition is 2/(x-2) + 5/(x-2)².
例子：分解 (2x+1)/(x-2)²。设 (2x+1)/(x-2)² = A/(x-2) + B/(x-2)²。去分母：2x+1 = A(x-2) + B。代入 x=2 得 5 = B。然后展开：2x+1 = Ax - 2A + 5。比较 x 系数得 2 = A。分解式为 2/(x-2) + 5/(x-2)²。
5. Irreducible Quadratic Factors | 不可约二次因式
When the denominator contains a quadratic factor that cannot be factorised into real linear factors (discriminant less than 0), the numerator in the partial fraction term must be of the form Ax+B, i.e., a linear polynomial. For example, for a denominator like (x²+1)(x-3), the decomposition is (Ax+B)/(x²+1) + C/(x-3).
当分母中含有无法分解为实一次因式的二次因式（判别式小于0）时，部分分式的该项分子必须为 Ax+B 的形式，即一个一次多项式。例如，对于分母 (x²+1)(x-3)，分解式为 (Ax+B)/(x²+1) + C/(x-3)。
The reason a constant numerator is insufficient is that over an irreducible quadratic, a linear expression is needed to represent any possible remainder when dividing a polynomial. After setting up the equation, multiply through and equate coefficients for x², x, and the constant term. You can also substitute values that simplify the equation, but since the quadratic has no real roots, equating coefficients is the primary method.
之所以仅使用常数分子不够，是因为在不可约二次式上，需要一个一次式才能表示多项式相除时可能出现的任何余式。建立方程后，通乘分母并比较 x²、x 及常数项的系数。也可以代入那些能使方程简化的数值，但由于该二次式没有实根，比较系数是主要方法。
: A+C=1, -A+B=2, -B+C=3. Solve to find A, B, C. The decomposition gives the required form.
例子：分解 (x²+2x+3)/((x²+1)(x-1))。设 = (Ax+B)/(x²+1) + C/(x-1)。通乘：x²+2x+3 = (Ax+B)(x-1) + C(x²+1)。展开：= Ax² - Ax + Bx - B + Cx² + C。合并：(A+C)x² + (-A+B)x + (-B+C) = x² + 2x + 3。比较系数：A+C=1, -A+B=2, -B+C=3。解方程组得出 A, B, C。分解式即得所需形式。
6. Repeated Quadratic Factors | 重复二次因式
For an irreducible quadratic factor raised to a power m, the partial fraction expansion includes a sum of terms (A₁x+B₁)/(quadratic) + (A₂x+B₂)/(quadratic)² + ... + (Aₘx+Bₘ)/(quadratic)ᵐ. As with repeated linear factors, omitting any intermediate power leads to an incomplete decomposition. The technique requires careful expansion and solving a system of equations.
对于不可约二次因式的 m 次幂，部分分式展开包含一串项：(A₁x+B₁)/(二次式) + (A₂x+B₂)/(二次式)² + …… + (Aₘx+Bₘ)/(二次式)ᵐ。同重复一次因式一样，遗漏任何一个中间次幂都会导致分解不完整。运用这一技巧需要仔细展开并求解方程组。
It is often efficient to combine substitution (if real linear factors also exist) and coefficient comparison. Substitute enough x values to generate simple equations, then use them alongside coefficient equating to reduce algebraic work. Consider factoring out common expressions to simplify.
将代入法（若同时也存在实一次因式）与系数比较法结合起来往往更加高效。代入足够多的 x 以产生简单方程，然后与系数比较结合使用，以减少代数运算量。适当提取公因式也会使过程简化。
pand and equate coefficients. While not common at SL, repeated quadratics appear in HL and further calculus contexts.
例子：(x)/((x²+4)²) 分解为 (Ax+B)/(x²+4) + (Cx+D)/((x²+4)²)。去分母：x = (Ax+B)(x²+4) + Cx+D。展开后比较系数。虽然在SL中不常见，但在HL和更深入的微积分情境中会出现重复二次因式。
7. The Cover-Up Rule | 遮盖法
re linear and non-repeated. To find the constant A corresponding to a linear factor (x-a), "cover up" . This works because multiplying by (x-a) and then setting x=a eliminates all other terms.
当所有因式均为无重复的一次因式时，遮盖法是求部分分式分子的快捷方法。要找到对应于一次因式 (x-a) 的常数 A，只需在原表达式中“遮盖”掉 (x-a) 这个因式，再将 x = a 代入剩下的表达式即可。这个方法的原理是：乘以 (x-a) 之后令 x=a，会消去所有其他项。
=-4: (-12+2)/(-4-1) = (-10)/(-5) = 2, so B=2. Thus the decomposition is 1/(x-1) + 2/(x+4).
例如，分解 (3x+2)/((x-1)(x+4))，要求 A/(x-1) 中的 A 时，遮盖 (x-1)，计算 (3x+2)/(x+4) 在 x=1 处的值：(3(1)+2)/(1+4) = 5/5 = 1，所以 A=1。求 B 时，遮盖 (x+4)，计算 (3x+2)/(x-1) 在 x=-4 处的值：(-12+2)/(-4-1) = (-10)/(-5) = 2，因此 B=2。于是分解式为 1/(x-1) + 2/(x+4)。
However, the cover-up rule does not directly apply to repeated or quadratic factors. In such cases, use it only for the highest power of a linear factor to find its coefficient, then use other methods for the remaining terms.
但遮盖法并不能直接用于重复因式或二次因式。在这些情况中，只可将其用于一次因式的最高次幂以求出相应系数，然后使用其他方法求取其余项。
8. Solving by Equating Coefficients | 比较系数法求解
When substitution alone cannot determine all constants (for example, when irreducible quadratics are present), equating coefficients becomes the main tool. After clearing denominators and expanding the right-hand side, you collect coefficients of like powers of x. Then set up a system of linear equations by matching the coefficients from the left-hand side polynomial. This method guarantees a solution provided the decomposition form is correct.
当仅靠代入法无法求出所有常数时（例如，存在不可约二次因式时），比较系数就成为主要手段。在去分母并展开右边之后，合并 x 的同次幂系数，再通过与左边多项式的系数进行匹配，从而建立线性方程组。只要分解形式正确，此方法必定能得出解。
/(x-2), multiply out and obtain an equation. Expanding yields terms in x³, x², x, and constant. By equating the coefficients of x³, x², x, and the constant, you generate a system. Solve using substitution or elimination. This systematic approach is robust and commonly required in IB exams.
例如，建立等式 (2x²+5)/((x²+1)(x-2)) = (Ax+B)/(x²+1) + C/(x-2) 后，乘开得到一个方程。展开后产生 x³, x², x 和常数项。通过比较 x³, x², x 的系数以及常数项，即生成方程组。用代入法或消元法求解。这一系统化方法十分可靠，在IB考试中经常需要用到。
Keep the algebra neat by working with one equation at a time and checking back-substitutions. If you obtain contradictory equations, it usually indicates a mistake in the assumed partial fraction form or an expansion error.
将代数运算处理得整洁有序，一次处理一个方程，并及时验算回代。如果得到矛盾的方程，通常意味着假定的部分分式形式有误或展开过程出现错误。
9. Integration Using Partial Fractions | 积分中的应用
) dx often become straightforward once the expression is decomposed. Each term integrates to a natural logarithm for linear denominators, or an arctangent and logarithm for quadratic denominators.
IB学生学习部分分式的主要原因之一就是为了对有理函数进行积分。形如 ∫ (多项式)/(多项式) dx 的积分，一旦表达式被分解，通常会变得简单易行。一次分母项积分后得到自然对数，二次分母项则可能得到反正切和对数。
After decomposition, integrate term by term: ∫ A/(x-a) dx = A ln|x-a| . This technique transforms seemingly impossible integrals into manageable standard forms.
分解后，逐项积分：∫ A/(x-a) dx = A ln|x-a| + C；∫ (Ax+B)/(x²+a²) dx 可通过代换拆成一个对数部分和一个反正切部分。这一方法将看似不可能的积分转变为可以操作的标准形式。
: (3x-1)/((x-2)(x+1)) = A/(x-2) + B/(x+1). Solving gives A and B; integrate to obtain A ln|x-2| + B ln|x+1| + C. Such problems are staples in IB HL papers.
例：计算 ∫ (3x-1)/(x²-x-2) dx。分解分母：(x-2)(x+1)。部分分式：(3x-1)/((x-2)(x+1)) = A/(x-2) + B/(x+1)。求解得出 A 和 B；积分后得到 A ln|x-2| + B ln|x+1| + C。此类问题是IB HL试卷中的常客。
10. Common Pitfalls and Tips | 常见误区与技巧
Students often forget to check whether the rational function is proper; starting partial fractions on an improper fraction without long division leads to a flawed equation. Always perform long division first if the numerator's degree is equal to or greater than the denominator's.
学生经常忘记检查有理函数是否为真分式；对假分式直接进行部分分式分解而不先做长除法，会导致方程错误。若分子次数等于或高于分母次数，务必先执行长除法。
Another error is omitting the required terms for repeated factors. If the denominator has (x+2)², the decomposition must include both A/(x+2) and B/(x+2)². Using only one of them results in an under-determined system or an identity mismatch.
另一个错误是遗漏重复因式所需的项。如果分母有 (x+2)²，分解式必须同时包含 A/(x+2) 和 B/(x+2)²。只使用其中一项会导致方程组欠定或恒等不匹配。
When dealing with quadratics, remember that the numerator is linear, not constant. Also, do not forget to integrate correctly: terms like 1/(x²+a²) integrate to (1/a) arctan(x/a), while x/(x²+a²) integrates to (1/2) ln(x²+a²) via substitution.
处理二次因式时，切记分子是一次式，而不是常数。另外，不要忘记正确积分：像 1/(x²+a²) 这样的项积分结果是 (1/a) arctan(x/a)，而 x/(x²+a²) 则通过代换积出 (1/2) ln(x²+a²)。
Practice systematically: write the correct form, clear denominators and label equations. For complex decompositions, use a mix of substitution and coefficient equating. Check your final decomposition by combining the partial fractions back to a single fraction to ensure they match the original expression.
要有条理地练习：写出正确的分解形式，去分母并给方程编号。对于复杂分解，可混合使用代入法与比较系数法。通过将部分分式重新合并成一个分式来检验最终分解结果，确保与原表达式一致。
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