📚 Exercise 16G: Integration by Partial Fractions | 练习16G：部分分式积分法
In IB Mathematics, particularly within the Analysis and Approaches HL syllabus, integrating rational functions is a fundamental skill. Exercise 16G focuses on the technique of partial fractions, which allows us to break down complex rational expressions into simpler ones that can be integrated directly using standard forms. Mastering this method is essential for tackling a wide range of calculus problems, from indefinite integrals to definite integrals and differential equations.
在IB数学中，尤其在分析与方法HL课程大纲里，有理函数的积分是一项基本技能。练习16G重点讲解部分分式法，该技巧能将复杂的有理式分解为更简单的分式，从而可以直接用标准积分公式求解。掌握这一方法对于应对从不定积分到定积分和微分方程的各类微积分问题至关重要。
1. Introduction to Partial Fractions | 部分分式简介
Partial fraction decomposition is a process of expressing a proper rational function P(x)/Q(x), where the degree of P is less than that of Q, as a sum of simpler fractions. Once decomposed, each term typically integrates into logarithmic or inverse tangent forms. The key is factoring the denominator Q(x) into irreducible polynomials: linear factors (ax + b) and quadratic factors (ax² + bx + c) that cannot be further factored over the real numbers.
部分分式分解是将一个真有理函数 P(x)/Q(x)（其中 P 的次数低于 Q 的次数）表示为若干简单分式之和的过程。一旦完成分解，每一项通常可直接积出对数或反正切形式的函数。关键在于将分母 Q(x) 分解为不可约多项式：线性因式 (ax + b) 和在实数范围内不能再分解的二次因式 (ax² + bx + c)。
2. Decomposing with Distinct Linear Factors | 不同线性因式的分解
When Q(x) factors into distinct linear terms (x - r₁)(x - r₂)...(x - rₙ), we set up the decomposition as A₁/(x - r₁) + A₂/(x - r₂) + ... + Aₙ/(x - rₙ). For example, for 1/((x-1)(x+3)) we write A/(x-1) + B/(x+3). Clearing denominators yields an equation that holds for all x, allowing us to find the constants A and B.
当 Q(x) 分解为不同的线性因式 (x - r₁)(x - r₂)...(x - rₙ) 时，我们将分解式设为 A₁/(x - r₁) + A₂/(x - r₂) + ... + Aₙ/(x - rₙ)。例如，对于 1/((x-1)(x+3))，我们写为 A/(x-1) + B/(x+3)。通过去分母得到一个对所有 x 成立的等式，从而求出常数 A 和 B。
3. Solving for Constants: Substitution Method | 求解常数：代入法
tes A, giving 1 = B(-4) so B = -1/4. This method is efficient for distinct linear factors.
为了快速求解 A 和 B，我们可以代入使各个线性因式为零的方便 x 值。在上例中，两边乘以 (x-1)(x+3) 得到 1 = A(x+3) + B(x-1)。令 x = 1 消去 B，得 1 = A(4)，故 A = 1/4。令 x = -3 消去 A，得 1 = B(-4)，故 B = -1/4。对于不同的线性因式，这一方法非常高效。
4. Repeated Linear Factors | 重复线性因式
When the denominator contains a repeated linear factor such as (x - r)ⁿ, the decomposition must include a series of terms: A₁/(x - r) + A₂/(x - r)² + ... + Aₙ/(x - r)ⁿ. For example, for (2x+1)/(x-2)², we write A/(x-2) + B/(x-2)². This ensures that all possible degrees of the repeated root are covered, and the resulting integrals involve both logarithms and negative power rules.
当分母包含如 (x - r)ⁿ 的重复线性因式时，分解式必须包含一系列项：A₁/(x - r) + A₂/(x - r)² + ... + Aₙ/(x - r)ⁿ。例如，对于 (2x+1)/(x-2)²，我们写为 A/(x-2) + B/(x-2)²。这确保了覆盖重复根的所有可能次数，并且积分结果既包含对数也包含负幂次法则。
+ C. Combining with the logarithm from the A/(x-2) term gives the complete antiderivative.
要积分像 B/(x-2)² 这样的项，回想 ∫ x⁻ⁿ dx = x⁻ⁿ⁺¹/(-n+1) + C。此处 ∫ (x-2)⁻² dx = -(x-2)⁻¹ + C。与来自 A/(x-2) 项的对数结合即可得到完整的原函数。
5. Integrating Logarithmic Forms | 对数形式的积分
The standard result ∫ 1/(ax+b) dx = (1/a) ln|ax+b| inator yields such a logarithmic term. For example, integrating (1/4)/(x-1) gives (1/4) ln|x-1|, and integrating (-1/4)/(x+3) gives (-1/4) ln|x+3|.
标准结果 ∫ 1/(ax+b) dx = (1/a) ln|ax+b| + C 是部分分式积分法的核心。一旦求出常数，每个线性分母的分式都会生成这样一个对数项。例如，积分 (1/4)/(x-1) 得到 (1/4) ln|x-1|，积分 (-1/4)/(x+3) 得到 (-1/4) ln|x+3|。
Here is a simple example in tabular form:
| Integral | Result |
|---|---|
| ∫ 1/((x-1)(x+3)) dx | (1/4) ln|x-1| - (1/4) ln|x+3| + C |
下表为一个简单示例：
| 积分 | 结果 |
|---|---|
| ∫ 1/((x-1)(x+3)) dx | (1/4) ln|x-1| - (1/4) ln|x+3| + C |
6. Irreducible Quadratic Factors | 不可约二次因式
If Q(x) has an irreducible quadratic factor ax² + bx + c (with discriminant Δ < 0), the corresponding partial fraction takes the form (Ax + B)/(ax² + bx + c). Integrating such a term often requires completing the square in the denominator and then splitting the numerator to match the derivative of the denominator, leading to a combination of ln and arctan forms.
如果 Q(x) 含有不可约二次因式 ax² + bx + c（判别式 Δ < 0），相应的部分分式形式为 (Ax + B)/(ax² + bx + c)。积分此类项通常需要将分母配方，然后拆分分子以匹配分母的导数，从而得到 ln 和 arctan 形式的组合。
integral splits into ∫ 2(x+1)/((x+1)²+4) dx - ∫ 1/((x+1)²+4) dx. The first part yields ln|(x+1)²+4| lds (1/2) arctan((x+1)/2) after applying the formula ∫ 1/(u² + a²) du = (1/a) arctan(u/a).
例如，考虑 ∫ (2x+1)/(x²+2x+5) dx。首先注意到 x²+2x+5 = (x+1)² + 4。将 2x+1 写为 2(x+1) - 1。于是积分拆分为 ∫ 2(x+1)/((x+1)²+4) dx - ∫ 1/((x+1)²+4) dx。第一部分结果为 ln|(x+1)²+4|，第二部分应用公式 ∫ 1/(u² + a²) du = (1/a) arctan(u/a) 后得到 (1/2) arctan((x+1)/2)。
7. Handling Improper Rational Functions | 处理假有理函数
If the degree of the numerator is greater than or equal to the degree of the denominator, the rational function is improper. Before applying partial fractions, we must perform polynomial long division to obtain a polynomial quotient plus a proper remainder over the denominator. The partial fraction decomposition is then applied only to the proper remainder part.
如果分子的次数大于或等于分母的次数，该有理函数为假分式。在应用部分分式之前，我们必须进行多项式长除法，得到一个多项式商加上一个真余式除以原分母。然后只对真余式部分使用部分分式分解。
sing partial fractions. The integral then becomes ∫ (x + 1/(x-1) + 1/(x+1)) dx = x²/2 + ln|x-1| + ln|x+1| + C.
例如，对于 ∫ (x³ + x)/(x² - 1) dx，用 x² - 1 除 x³ + x 得到 x + (2x)/(x² - 1)。余式 (2x)/(x² - 1) 可通过部分分式分解为 1/(x-1) + 1/(x+1)。于是积分变为 ∫ (x + 1/(x-1) + 1/(x+1)) dx = x²/2 + ln|x-1| + ln|x+1| + C。
8. Systematic Approach to Setting Up Decomposition | 建立分解式的系统方法
A structured procedure ensures accuracy: (1) Factor the denominator completely into linear and irreducible quadratic factors. (2) For each distinct linear factor (ax+b)¹, assign a term A/(ax+b). (3) For each repeated linear factor (ax+b)ⁿ, include terms A₁/(ax+b) + A₂/(ax+b)² + ... + Aₙ/(ax+b)ⁿ. (4) For each distinct irreducible quadratic factor (ax²+bx+c), assign (Ax+B)/(ax²+bx+c). (5) For repeated quadratics, follow a similar pattern with increasing powers.
一个系统的步骤可确保准确性：(1) 将分母完全分解为线性和不可约二次因式。(2) 对于每个不同的线性因式 (ax+b)¹，分配一项 A/(ax+b)。(3) 对于每个重复线性因式 (ax+b)ⁿ，包含 A₁/(ax+b) + A₂/(ax+b)² + ... + Aₙ/(ax+b)ⁿ。(4) 对于每个不同的不可约二次因式 (ax²+bx+c)，分配 (Ax+B)/(ax²+bx+c)。(5) 对于重复二次因式，按幂次递增的类似模式处理。
This framework prevents missing terms and guarantees that the decomposition is mathematically valid. Always check that the number of unknown constants equals the degree of the denominator. After solving for constants, verify by combining the partial fractions back to see if they match the original rational function.
这一框架可防止遗漏项，确保分解在数学上有效。务必检查未知常数的个数是否等于分母的次数。解出常数后，可通过合并部分分式来验证它们是否等于原来的有理函数。
9. Definite Integrals with Partial Fractions | 部分分式法的定积分
r example, ∫ from 0 to 1 of 1/((x+1)(x+2)) dx decomposes to ∫ (1/(x+1) - 1/(x+2)) dx = [ln|x+1| - ln|x+2|] from 0 to 1 = ln(2/3) - ln(1/2) = ln(4/3).
在计算定积分时，首先用部分分式法求出不定积分，然后按标准方法代入上下限。注意对数项：在对数内使用绝对值并尽可能合并。例如，∫₀¹ 1/((x+1)(x+2)) dx 分解为 ∫ (1/(x+1) - 1/(x+2)) dx = [ln|x+1| - ln|x+2|]₀¹ = ln(2/3) - ln(1/2) = ln(4/3)。
A quick reference for definite integrals with partial fractions:
| Step | 英文说明 | 中文说明 |
|---|---|---|
| 1 | Decompose the integrand | 分解被积函数 |
| 2 | Integrate each term | 逐项积分 |
| 3 | Apply limits carefully | 仔细代入上下限 |
定积分中使用部分分式的快速参考：
10. Common Pitfalls and How to Avoid Them | 常见误区及规避方法
One frequent mistake is forgetting to include enough terms for repeated factors. For example, for (x+1)/(x(x-1)²), the correct setup is A/x + B/(x-1) + C/(x-1)², not just A/x + C/(x-1)². Another error is misidentifying irreducible quadratics; always check the discriminant. Also, when solving for constants, relying solely on substitution may miss constants if not all factors are distinct; then equate coefficients of powers of x as a backup.
一个常见错误是对于重复因式忘记包含足够多的项。例如，对于 (x+1)/(x(x-1)²)，正确的设定是 A/x + B/(x-1) + C/(x-1)²，而不仅仅是 A/x + C/(x-1)²。另一个错误是误判不可约二次因式；务必检查判别式。此外，在求解常数时，如果并非所有因式都不同，单纯依赖代入法可能会遗漏常数；此时可使用比较同次幂系数的方法作为辅助。
To avoid algebraic slip-ups, always multiply both sides by the full denominator, expand, and group like terms. This allows you to form a system of linear equations for the constants. Double-check your final decomposition by putting it over a common denominator mentally or via simplification.
为避免代数失误，一定要将两边乘以完整分母，展开并合并同类项。这样可以为常数建立线性方程组。通过心算或化简将最终分解式通分来双重检验。
11. IB Exam-Style Question Breakdown | IB考试风格题目解析
A typical exam question might ask: "Find ∫ (3x² + 5x - 2)/(x³ - x² - x + 1) dx." ominator: x³ - x² - x + 1 = (x-1)²(x+1). Since the numerator degree (2) is less than 3, it' lve to get A = 2, B = 3, C = 1. Then integrate: ∫ (2/(x-1) + 3/(x-1)² + 1/(x+1)) dx = 2 ln|x-1| - 3/(x-1) + ln|x+1| + C.
一道典型考题可能要求：“求 ∫ (3x² + 5x - 2)/(x³ - x² - x + 1) dx。”首先分解分母：x³ - x² - x + 1 = (x-1)²(x+1)。由于分子次数(2)小于3，其为真分式。建立分解式：(3x²+5x-2)/((x-1)²(x+1)) = A/(x-1) + B/(x-1)² + C/(x+1)。求解得 A=2, B=3, C=1。然后积分：∫ (2/(x-1) + 3/(x-1)² + 1/(x+1)) dx = 2 ln|x-1| - 3/(x-1) + ln|x+1| + C。
Such questions often award marks for correct decomposition, solving for constants, and integrating each term accurately. Showing all steps clearly is vital. Be prepared to handle definite integrals as well by substituting limits and simplifying logarithmic expressions.
此类题目通常根据正确的分解、常数的求解和各项积分的准确性来给分。清晰展示所有步骤至关重要。同时要准备好处理定积分，通过代入上下限并化简对数表达式来求解。
12. Practice Summary and Key Takeaways | 练习总结与核心要点
Exercise 16G solidifies the integration of rational expressions via partial fractions, a skill that bridges algebra and calculus. The core takeaways include: (1) Always start by ensuring the rational function is proper; if not, perform polynomial division. (2) Factor the denominator completely and write the appropriate decomposition based on the nature of the factors. (3) Efficiently solve for constants using substitution and equating coefficients. (4) Integrate using fundamental logarithmic, power, and arctangent forms. Mastery here will support more advanced topics like differential equations, volume of revolution, and complex analysis in further mathematics.
练习16G巩固了通过部分分式积分有理表达式的技能，这是连接代数与微积分的桥梁。核心要点包括：(1) 始终先确保有理函数为真分式；若不是，则先进行多项式除法。(2) 完全分解分母，并根据因式的特性写出相应的分解式。(3) 利用代入法和比较系数法高效求解常数。(4) 使用基本的对数、幂函数和反正切形式进行积分。掌握此部分内容将为微分方程、旋转体积以及进阶数学中的复分析等更高级的主题提供支持。
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