📚 B8I Part 6: Partial Fractions in Integration | B8I 第六部分：部分分式积分法
In A-level Mathematics, many rational functions cannot be integrated directly using standard results. The method of partial fractions decomposes a complicated algebraic fraction into a sum of simpler fractions, each of which can then be integrated using basic log or arctan results.
在 A-level 数学中，许多有理函数无法直接使用标准公式进行积分。部分分式法将一个复杂的代数分式分解为若干个简单分式之和，然后可以逐项使用对数或反正切等基本结果进行积分。
1. When to Use Partial Fractions in Integration | 何时在积分中使用部分分式
Use partial fractions when you need to integrate a rational function whose denominator can be factorised, and the degree of the numerator is less than the degree of the denominator.
当你需要积分一个分母可以因式分解、且分子的次数低于分母次数的有理函数时，就应当使用部分分式法。
For example, ∫ (3x + 5)/((x – 1)(x + 2)) dx is a suitable candidate, but ∫ x/(x² + 1) dx is not, because substitution u = x² + 1 gives a faster result.
例如，∫ (3x + 5)/((x – 1)(x + 2)) dx 是适合使用部分分式的对象，但 ∫ x/(x² + 1) dx 则不适合，因为用换元 u = x² + 1 会更快。
2. Checking the Degree Condition | 检查次数条件
Before applying partial fractions, always check that the numerator has a lower degree than the denominator. This is called a proper rational function.
在应用部分分式之前，始终要检查分子的次数是否低于分母的次数。这样的函数称为真有理函数。
If the degree of the numerator is greater than or equal to the degree of the denominator, perform polynomial division first. The remainder term will then be a proper fraction and can be decomposed normally.
如果分子的次数大于或等于分母的次数，首先要进行多项式除法。得到的余项将是真分式，可以正常进行分解。
For example, (x³ + 2x + 1)/(x² – 1) must be divided first because the numerator degree 3 is greater than the denominator degree 2.
例如，(x³ + 2x + 1)/(x² – 1) 必须先做多项式除法，因为分子次数 3 大于分母次数 2。
3. Distinct Linear Factors | 不同的一次因式
If the denominator factorises into distinct linear factors such as (x – a)(x – b), write the fraction as A/(x – a) + B/(x – b).
如果分母分解为不同的一次因式，例如 (x – a)(x – b)，则将分式写成 A/(x – a) + B/(x – b)。
Multiply both sides by the full denominator and then substitute convenient values of x, such as x = a and x = b, to find the constants A and B quickly.
将等式两边同时乘以完整分母，然后代入方便的 x 值，例如 x = a 和 x = b，即可快速求出常数 A 和 B。
The resulting integrals have the form ∫ 1/(x – a) dx = ln|x – a| + C.
得到的积分形式为 ∫ 1/(x – a) dx = ln|x – a| + C。
4. Repeated Linear Factors | 重复的一次因式
If a linear factor appears more than once, such as (x – a)², you must include two terms: A/(x – a) + B/(x – a)².
如果某个一次因式出现多次，例如 (x – a)²，你必须包含两项：A/(x – a) + B/(x – a)²。
This is a common source of error in exams. Missing the repeated term leads to an incorrect decomposition and an incorrect integral.
这是考试中常见的错误来源。遗漏重复项会导致分解错误，进而导致积分错误。
For example, (4x + 1)/(x – 2)² becomes A/(x – 2) + B/(x – 2)², and the second term integrates to -B/(x – 2) + C.
例如，(4x + 1)/(x – 2)² 应分解为 A/(x – 2) + B/(x – 2)²，其中第二项积分为 -B/(x – 2) + C。
5. Irreducible Quadratic Factors | 不可约的二次因式
When the denominator contains a quadratic factor such as x² + 1 that cannot be factorised into real linear factors, the corresponding numerator must be of the form Ax + B.
当分母含有像 x² + 1 这样无法分解为实系数一次因式的二次因式时，对应的分子必须写成 Ax + B 的形式。
This allows you to split the resulting integral into a logarithm part and an arctangent part. For example, ∫ (Ax + B)/(x² + a²) dx leads to a combination of ln|x² + a²| and (1/a) arctan(x/a).
这样可以将得到的积分拆分为对数部分和反正切部分。例如，∫ (Ax + B)/(x² + a²) dx 会得到 ln|x² + a²| 和 (1/a) arctan(x/a) 的组合。
Completing the square is often needed if the quadratic factor is written in the form x² + bx + c.
如果二次因式写成 x² + bx + c 的形式，通常需要先配方。
6. Improper Fractions and Polynomial Division | 假分式与多项式除法
When the rational function is improper, divide the numerator by the denominator using polynomial long division. The quotient becomes a polynomial, and the remainder is a proper fraction.
当有理函数是假分式时，使用多项式长除法将分子除以分母。商式是一个多项式，余式是一个真分式。
For example, (x³ + 2)/(x² – 1) = x + (x + 2)/(x² – 1). The polynomial x can be integrated directly, and the remaining proper fraction can be handled by partial fractions.
例如，(x³ + 2)/(x² – 1) = x + (x + 2)/(x² – 1)。多项式 x 可以直接积分，剩下的真分式可以用部分分式处理。
Never apply partial fractions directly to an improper fraction; this wastes time and often produces incorrect constants.
切勿直接对假分式使用部分分式；这既浪费时间，又常常得到错误的常数。
7. Worked Example 1: Distinct Linear Factors | 例题一：不同的一次因式
Integrate ∫ (2x + 3)/((x – 1)(x + 2)) dx. First write the decomposition as A/(x – 1) + B/(x + 2).
求 ∫ (2x + 3)/((x – 1)(x + 2)) dx。首先写出分解式 A/(x – 1) + B/(x + 2)。
Multiplying through by (x – 1)(x + 2) gives 2x + 3 = A(x + 2) + B(x – 1). Substituting x = 1 gives 5 = 3A, so A = 5/3. Substituting x = -2 gives -1 = -3B, so B = 1/3.
两边同时乘以 (x – 1)(x + 2) 得 2x + 3 = A(x + 2) + B(x – 1)。代入 x = 1 得 5 = 3A，所以 A = 5/3。代入 x = -2 得 -1 = -3B，所以 B = 1/3。
Therefore the integral becomes (5/3)∫ 1/(x – 1) dx + (1/3)∫ 1/(x + 2) dx = (5/3)ln|x – 1| + (1/3)ln|x + 2| + C.
因此积分变为 (5/3)∫ 1/(x – 1) dx + (1/3)∫ 1/(x + 2) dx = (5/3)ln|x – 1| + (1/3)ln|x + 2| + C。
8. Worked Example 2: Repeated Linear Factor | 例题二：重复的一次因式
Integrate ∫ (3x + 5)/(x – 1)² dx. Write the decomposition as A/(x – 1) + B/(x – 1)².
求 ∫ (3x + 5)/(x – 1)² dx。写出分解式 A/(x – 1) + B/(x – 1)²。
Multiplying by (x – 1)² gives 3x + 5 = A(x – 1) + B. Expanding and comparing coefficients gives A = 3 and B = 8.
两边同时乘以 (x – 1)² 得 3x + 5 = A(x – 1) + B。展开并比较系数，得到 A = 3 和 B = 8。
The integral is then 3∫ 1/(x – 1) dx + 8∫ 1/(x – 1)² dx = 3ln|x – 1| – 8/(x – 1) + C.
积分变为 3∫ 1/(x – 1) dx + 8∫ 1/(x – 1)² dx = 3ln|x – 1| – 8/(x – 1) + C。
9. Worked Example 3: Irreducible Quadratic Factor | 例题三：不可约二次因式
Integrate ∫ (x + 1)/(x² + 2x + 5) dx. The denominator is irreducible, so complete the square first: x² + 2x + 5 = (x + 1)² + 4.
求 ∫ (x + 1)/(x² + 2x + 5) dx。分母不可约，因此先配方：x² + 2x + 5 = (x + 1)² + 4。
Let u = x + 1, then du = dx and the numerator x + 1 = u. The integral becomes ∫ u/(u² + 4) du.
令 u = x + 1，则 du = dx，分子 x + 1 = u。积分变为 ∫ u/(u² + 4) du。
This splits naturally as ½ ∫ 2u/(u² + 4) du, which integrates to ½ ln|u² + 4| + C = ½ ln|x² + 2x + 5| + C.
这自然拆分为 ½ ∫ 2u/(u² + 4) du，积分结果为 ½ ln|u² + 4| + C = ½ ln|x² + 2x + 5| + C。
If the numerator also had a constant term, the result would include an arctan term such as (1/2) arctan((x + 1)/2).
如果分子还含有常数项，结果中会出现反正切项，例如 (1/2) arctan((x + 1)/2)。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
A frequent mistake is forgetting the repeated term in decompositions such as (x + 1)/(x – 2)² by writing only A/(x – 2). This fails because two constants are required for a repeated linear factor.
一个常见错误是在 (x + 1)/(x – 2)² 这样的分解中只写 A/(x – 2)，遗漏了重复项。这是错误的，因为重复的一次因式需要两个待定常数。
Another trap is using partial fractions on a rational function like x/(x² + 1) when substitution is much quicker. Always check if the numerator is a multiple of the derivative of the denominator before decomposing.
另一个陷阱是对像 x/(x² + 1) 这样的有理函数使用部分分式，而实际上换元法要快得多。在分解之前，始终要检查分子是否为分母导数的倍数。
Finally, remember to include the absolute value inside logarithms when integrating fractions, and write + C for every indefinite integral.
最后，在对分式积分时记得在 ln 内加绝对值，并且每一个不定积分都要写 + C。
11. Summary of Decomposition Forms | 分解形式总结
| Denominator factor | Form of partial fractions | Typical integral result |
|---|---|---|
| x – a | A/(x – a) | A ln|x – a| + C |
| (x – a)² | A/(x – a) + B/(x – a)² | A ln|x – a| – B/(x – a) + C |
| x² + a² | (Ax + B)/(x² + a²) | (A/2)ln|x² + a²| + (B/a) arctan(x/a) + C |
Use this table as a quick reference when setting up decomposition before integration.
在设置积分前的分解式时，可以将此表作为快速参考。
12. Final Practice Advice | 最后练习建议
Begin with simple distinct linear factors to build confidence, then move to repeated roots and quadratic factors. Always verify your decomposition by combining the right-hand side back to the original fraction.
先从简单的不同一次因式开始练习以建立信心，然后再处理重根和二次因式。始终通过将右边合并回原分式来检验你的分解是否正确。
In timed exams, spend the first 30 seconds checking the degree condition and looking for a faster substitution. Partial fractions are powerful but should be used only when they are the shortest route to the integral.
在限时考试中，花前 30 秒检查次数条件并寻找是否有更快的换元法。部分分式虽然强大，但只应在它是积分最短路径时才使用。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导