📚 Partial Fractions | 部分分式
Partial fractions are a standard A-level technique for rewriting a single rational function as a sum of simpler fractions. They are especially useful for integration, binomial expansion, and solving differential equations.
部分分式是 A-level 的标准技巧，用于把一个有理函数改写为几个更简单分式之和。它在积分、二项展开和求解微分方程中特别有用。
1. What Are Partial Fractions? | 什么是部分分式？
A rational function is the quotient of two polynomials. For example, f(x) = (2x + 3) / ((x + 1)(x – 2)) is a rational function. Partial fractions express it as A/(x + 1) + B/(x – 2), where A and B are constants to be found.
有理函数是两个多项式之商。例如 f(x) = (2x + 3) / ((x + 1)(x – 2)) 就是一个有理函数。部分分式将它表示为 A/(x + 1) + B/(x – 2)，其中 A 和 B 是待求常数。
The method works only when the denominator can be factorised into linear or quadratic factors over the real numbers. The numerator must have degree less than the denominator unless polynomial division is used first.
只有当分母可以在实数范围内分解为一次或二次因式时，该方法才适用。分子次数必须低于分母次数，除非先进行多项式除法。
2. Proper and Improper Rational Functions | 真分式与假分式
A proper rational function has numerator degree strictly less than denominator degree. An improper rational function has numerator degree greater than or equal to denominator degree.
真分式是指分子的次数严格小于分母的次数。假分式是指分子的次数大于或等于分母的次数。
For an improper fraction such as (x³ + 2x + 1)/(x² – 1), you must divide first to get a polynomial plus a proper remainder. Only then can you decompose the remainder into partial fractions.
对于像 (x³ + 2x + 1)/(x² – 1) 这样的假分式，必须先做除法，得到一个多项式加上一个真余式。然后才能对余式进行部分分式分解。
In Edexcel questions, the improper case is common. Forgetting to divide first leads to an incorrect assumed form, because the numerator would effectively have too high a degree.
在 Edexcel 考试中，假分式的情形很常见。忘记先做除法会导致假设形式错误，因为分子次数实际上过高。
3. Distinct Linear Factors | 不同的一次因式
If the denominator factors into distinct linear factors, write one term for each factor. For example, 3/((x – 1)(x + 2)) = A/(x – 1) + B/(x + 2).
如果分母分解为不同的一次因式，则每个因式对应一项。例如 3/((x – 1)(x + 2)) = A/(x – 1) + B/(x + 2)。
Multiply both sides by the original denominator to clear fractions. Then choose values of x that make each linear factor zero to find A and B quickly.
两边同乘以原分母以消去分式。然后选择使每个一次因式为零的 x 值，快速求出 A 和 B。
Worked example: 3/((x – 1)(x + 2)) = A/(x – 1) + B/(x + 2). Multiply by (x – 1)(x + 2): 3 = A(x + 2) + B(x – 1). Let x = 1: 3 = 3A, so A = 1. Let x = -2: 3 = -3B, so B = -1. Thus 3/((x – 1)(x + 2)) = 1/(x – 1) – 1/(x + 2).
例题：3/((x – 1)(x + 2)) = A/(x – 1) + B/(x + 2)。两边乘以 (x – 1)(x + 2)：3 = A(x + 2) + B(x – 1)。令 x = 1：3 = 3A，所以 A = 1。令 x = -2：3 = -3B，所以 B = -1。因此 3/((x – 1)(x + 2)) = 1/(x – 1) – 1/(x + 2)。
4. Repeated Linear Factors | 重复的一次因式
If a linear factor is repeated, include all powers from 1 up to the repeated power. For a denominator (x + 1)²(x – 3), write A/(x + 1) + B/(x + 1)² + C/(x – 3).
如果一次因式重复出现，需包含从 1 次幂到重复次幂的所有项。例如分母为 (x + 1)²(x – 3)，写成 A/(x + 1) + B/(x + 1)² + C/(x – 3)。
Do not write only A/(x + 1)² + C/(x – 3); the term A/(x + 1) is also needed to represent all possible numerators after combining over the common denominator.
不要只写 A/(x + 1)² + C/(x – 3)；还需要 A/(x + 1) 这一项，以覆盖通分后所有可能的分子。
Example: (5x + 7)/((x + 1)(x + 2)²) = A/(x + 1) + B/(x + 2) + C/(x + 2)². Multiplying gives 5x + 7 = A(x + 2)² + B(x + 1)(x + 2) + C(x + 1). Substitute x = -2: -10 + 7 = C(-1), so C = 3. Substitute x = -1: -5 + 7 = A(1), so A = 2. Then compare coefficients of x² or substitute another value to find B = -2.
例如：(5x + 7)/((x + 1)(x + 2)²) = A/(x + 1) + B/(x + 2) + C/(x + 2)²。通分得 5x + 7 = A(x + 2)² + B(x + 1)(x + 2) + C(x + 1)。代入 x = -2：-10 + 7 = C(-1)，所以 C = 3。代入 x = -1：-5 + 7 = A(1)，所以 A = 2。然后比较 x² 的系数或代入另一个值求出 B = -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, such as x² + 1 or x² + 2x + 5, the numerator for that term must be linear.
当分母含有无法在实数范围内分解为一次因式的二次因式时，例如 x² + 1 或 x² + 2x + 5，该项的分子必须是一次式。
Write (Ax + B)/(x² + 1), not just A/(x² + 1). This is because a linear numerator is needed to match all possible remainders after division by a quadratic denominator.
应写成 (Ax + B)/(x² + 1)，而不是仅写 A/(x² + 1)。这是因为除以二次分母后，余式可能是一次式，需要一次分子来匹配。
Example: (3x + 5)/((x – 1)(x² + 4)) = A/(x – 1) + (Bx + C)/(x² + 4). Multiply: 3x + 5 = A(x² + 4) + (Bx + C)(x – 1). Let x = 1: 8 = 5A, so A = 8/5. Then expand and equate coefficients to find B = -8/5 and C = -7/5.
例如：(3x + 5)/((x – 1)(x² + 4)) = A/(x – 1) + (Bx + C)/(x² + 4)。通分：3x + 5 = A(x² + 4) + (Bx + C)(x – 1)。令 x = 1：8 = 5A，所以 A = 8/5。然后展开并比较系数，求出 B = -8/5，C = -7/5。
6. Polynomial Division Before Decomposition | 分解前的多项式除法
If the rational function is improper, perform algebraic long division first. The result is a polynomial quotient plus a proper rational remainder.
如果有理函数是假分式，首先进行代数长除法。结果是一个多项式商加上一个真分式余式。
For example, (x³ + 2x² + x + 1)/(x² + 1). Long division gives x + 2 + (-x – 1)/(x² + 1). Then decompose the proper remainder if possible.
例如 (x³ + 2x² + x + 1)/(x² + 1)。长除法得到 x + 2 + (-x – 1)/(x² + 1)。然后如有可能，对真分式余式进行分解。
In many exam questions, the improper part is given so that you first divide, then apply partial fractions to the remainder. Forgetting this step is a common mistake that changes the whole problem.
在许多考试题中，会给出假分式部分，你需要先做除法，再对余式应用部分分式。忘记这一步是常见错误，会改变整个题目。
7. Equating Coefficients | 比较系数法
After clearing fractions, you can expand the right-hand side and compare coefficients of powers of x. This is useful when substituting roots is not enough or when fractions are complex.
消去分式后，可以展开右边并比较 x 的各次幂系数。当代入根值不够用或分式较复杂时，这一方法很有用。
For example, suppose 2x² + 3x + 1 = A(x² + 1) + (Bx + C)(x – 2). Expand to get A x² + A + B x² – 2B x + C x – 2C. Compare coefficients of x², x, and constant to form simultaneous equations.
例如，假设 2x² + 3x + 1 = A(x² + 1) + (Bx + C)(x – 2)。展开得到 A x² + A + B x² – 2B x + C x – 2C。比较 x²、x 和常数项的系数，建立联立方程组。
This method is systematic and less dependent on choosing ‘nice’ values of x. However, it can be slower than substitution for distinct linear factors.
该方法系统性强，不太依赖选择“好”的 x 值。但对于不同的一次因式，它可能比代入法慢。
8. Integration Using Partial Fractions | 利用部分分式积分
One of the most common applications is integrating rational functions. After decomposition, each term integrates to a natural logarithm or inverse tangent, depending on the denominator.
最常见的应用之一是对有理函数积分。分解后，每一项积分为自然对数或反正切，具体取决于分母。
For linear denominators: ∫ A/(ax + b) dx = (A/a) ln|ax + b| + C. For quadratic denominators like x² + a², use ∫ (Bx)/(x² + a²) dx = (B/2) ln(x² + a²) + C and ∫ C/(x² + a²) dx = (C/a) arctan(x/a) + C.
对于一次分母：∫ A/(ax + b) dx = (A/a) ln|ax + b| + C。对于像 x² + a² 这样的二次分母，使用 ∫ (Bx)/(x² + a²) dx = (B/2) ln(x² + a²) + C 以及 ∫ C/(x² + a²) dx = (C/a) arctan(x/a) + C。
Example: ∫ (1/(x – 1) – 1/(x + 2)) dx = ln|x – 1| – ln|x + 2| + C = ln| (x – 1)/(x + 2) | + C.
例如：∫ (1/(x – 1) – 1/(x + 2)) dx = ln|x – 1| – ln|x + 2| + C
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