📚 pdfjoiner_(4)-038: Integration by Partial Fractions | pdfjoiner_(4)-038：部分分式积分法
ppears in exam papers. The worksheet labelled pdfjoiner_(4)-038 contains a classic problem that requires decomposing an algebraic fraction into simpler parts before integration. Mastering this method—integration by partial fractions— for tackling a whole family of integrals that standard techniques cannot handle directly.
在A-Level数学中，尤其是在Edexcel纯数学课程中，积分有理函数是试卷中经常出现的技能。这份名为pdfjoiner_(4)-038的练习卷中有一道经典题目，要求先将代数分式分解为更简单的部分再进行积分。掌握这种方法——部分分式积分法——将赋予你一个强大的工具，用以应对一系列标准技巧无法直接处理的积分。
1. When Do You Need Partial Fractions for Integration? | 什么时候积分需要部分分式？
If you are asked to integrate a rational expression of the form P(x)/Q(x) where P and Q are polynomials, and the denominator Q(x) can be factorised into linear or quadratic factors, writing the expression as a sum of simpler fractions often makes integration straightforward. You will see this technique applied when the integrand is not immediately recognisable as a standard derivative.
如果你被要求对形如P(x)/Q(x)的有理式积分，其中P与Q是多项式，且分母Q(x)可以分解为线性或二次因式，那么将该式写成几个更简单分式的和通常会让积分变得直接。当被积函数不能立即识别为标准导数的形式时，你便会看到该技巧的应用。
2. The Core Idea: Decompose into Simpler Pieces | 核心思想：拆解为更简单的部分
The method relies on the fact that any proper rational function (degree of numerator less than degree of denominator) can be expressed as a sum of partial fractions. Each partial fraction has a denominator that is a factor of Q(x), and its numerator is a constant (or a linear term when the factor is an irreducible quadratic). Once decomposed, the integral breaks into a sum of basic integrals, typically leading to natural logarithms or inverse tangent functions.
该方法基于这样一个事实：任何真分式（分子次数低于分母次数）都可以表示为部分分式的和。每个部分分式的分母都是Q(x)的一个因式，而其分子是一个常数（当因式为不可约二次式时，分子可能是一次式）。分解之后，积分就变成若干个基本积分的和，通常会得到自然对数或反正切函数。
3. Case 1: Distinct Linear Factors | 情形一：不同线性因式
+ d), the partial fraction decomposition takes the form A/(ax+b) + B/(cx+d). The constants A and B are found by multiplying through by the denominator and comparing coefficients or substituting convenient values of x. Integrating each term yields A/a ln|ax+b| + B/c ln|cx+d| + C.
当Q(x)分解为不同的线性因式，例如Q(x) = (ax + b)(cx + d)，部分分式的形式为A/(ax+b) + B/(cx+d)。常数A和B可以通过两边乘以分母并对比系数或代入便利的x值来求出。逐项积分得到 A/a ln|ax+b| + B/c ln|cx+d| + C。
4. Worked Example – Distinct Linear Factors | 例题 – 不同线性因式
lying both sides by (x+1)(x+2) gives 2x+3 = A(x+2) + B(x+1). Setting x = -1 gives 1 = A(1) ⇒ A = 1. Setting x = -2 gives -1 = B(-1) ⇒ B = 1. So the integrand becomes 1/(x+1) + 1/(x+2). Integrating gives ln|x+1| + ln|x+2| + C.
考虑积分 ∫ (2x+3)/(x² + 3x + 2) dx。分母分解为 (x+1)(x+2)。将分式写作 A/(x+1) + B/(x+2)。两边乘以 (x+1)(x+2) 得到 2x+3 = A(x+2) + B(x+1)。令 x = -1 得 1 = A(1) ⇒ A = 1。令 x = -2 得 -1 = B(-1) ⇒ B = 1。因此被积函数变为 1/(x+1) + 1/(x+2)。积分得到 ln|x+1| + ln|x+2| + C。
5. Case 2: Repeated Linear Factors | 情形二：重复线性因式
If Q(x) contains a repeated linear factor such as (ax + b)², the partial fraction expansion includes terms with that factor to powers 1 and 2: A/(ax+b) + B/(ax+b)². The same logic extends to higher powers. Integration of the B/(ax+b)² term uses the power rule after substitution, giving -B/[a(ax+b)].
若Q(x)含有重复线性因式，如 (ax + b)²，则部分分式展开必须包含该因式的一次和二次项：A/(ax+b) + B/(ax+b)²。相同的逻辑可推广至更高次幂。对 B/(ax+b)² 项的积分通过换元后应用幂法则，得到 -B/[a(ax+b)]。
6. Worked Example – Repeated Factor | 例题 – 重复因式
A(x-1) + B. Compare coefficients or substitute x=1: 4 = B. Then expanded: x+3 = Ax - A + 4 ⇒ Ax - A = x - 1 ⇒ A = 1. The integral becomes ∫ [1/(x-1) + 4/(x-1)²] dx = ln|x-1| - 4/(x-1) + C.
求 ∫ (x+3)/(x² - 2x + 1) dx。分母为 (x-1)²，故设 (x+3)/(x-1)² = A/(x-1) + B/(x-1)²。两边乘开：x+3 = A(x-1) + B。对比系数或代入 x=1：4 = B。展开得 x+3 = Ax - A + 4 ⇒ Ax - A = x - 1 ⇒ A = 1。积分变为 ∫ [1/(x-1) + 4/(x-1)²] dx = ln|x-1| - 4/(x-1) + C。
7. Handling Improper Algebraic Fractions | 处理假分式
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, you must perform polynomial long division to express it as a polynomial plus a proper fraction. Only the proper part is decomposed further. The polynomial part integrates directly.
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，必须先进行多项式长除法，将其表示为一个多项式加上一个真分式。只有真分式部分需要继续分解。多项式部分可以直接积分。
8. Full Walkthrough of a pdfjoiner_(4)-038 Style Problem | pdfjoiner_(4)-038 类型题目的完整剖析
inator, so we start with division. Divide 2x²+5x+1 by x²+3x+2. The leading term gives quotient 2. Multiply: 2(x²+3x+2) = 2x²+6x+4. Subtract from the numerator: (2x²+5x+1) - (2x²+6x+4) = -x -3. Hence the integrand rewrites as 2 + (-x -3)/(x²+3x+2).
练习卷 pdfjoiner_(4)-038 中的问题要求我们计算 ∫ (2x² + 5x + 1)/(x² + 3x + 2) dx。注意到分子与分母次数相同，因此我们从除法开始。用 x²+3x+2 除 2x²+5x+1。首项商为 2。相乘：2(x²+3x+2) = 2x²+6x+4。从分子中减去：(2x²+5x+1) - (2x²+6x+4) = -x -3。因此被积函数重写为 2 + (-x -3)/(x²+3x+2)。
/(x+1) + B/(x+2). Multiply to get -x -3 = A(x+2) + B(x+1). Let x = -1: -(-1)-3 = -2 = A(1) ⇒ A = -2. Let x = -2: -(-2)-3 = -1 = B(-1) ⇒ B = 1. The proper fraction becomes -2/(x+1) + 1/(x+2).
现在对分母因式分解：x²+3x+2 = (x+1)(x+2)。设 (-x -3)/[(x+1)(x+2)] = A/(x+1) + B/(x+2)。通分得 -x -3 = A(x+2) + B(x+1)。令 x = -1：-(-1)-3 = -2 = A(1) ⇒ A = -2。令 x = -2：-(-2)-3 = -1 = B(-1) ⇒ B = 1。真分式变为 -2/(x+1) + 1/(x+2)。
∫ (2x²+5x+1)/(x²+3x+2) dx = ∫ [2 - 2/(x+1) + 1/(x+2)] dx
Integrating term by term: ∫2 dx = 2x, ∫ -2/(x+1) dx = -2 ln|x+1|, ∫ 1/(x+2) dx = ln|x+2|. Therefore the final answer is 2x - 2 ln|x+1| + ln|x+2| + C.
逐项积分：∫2 dx = 2x，∫ -2/(x+1) dx = -2 ln|x+1|，∫ 1/(x+2) dx = ln|x+2|。因此最终答案为 2x - 2 ln|x+1| + ln|x+2| + C。
9. Definite Integration with Partial Fractions | 定积分背景下使用部分分式
When the problem includes limits, the procedure is identical, but you must remember to apply the limits after integrating and simplify the logarithmic expressions carefully. Combining logarithms using laws of logs can often give a neater exact value. Always check for domain issues: the denominator must not be zero on the interval of integration.
当题目包含上下限时，步骤完全相同，但必须记住在积分之后代入上下限并小心化简对数式。利用对数律合并对数通常能给出更简洁的精确值。务必检查定义域问题：在积分区间上分母不得为零。
10. Common Mistakes and How to Avoid Them | 常见错误与规避方法
setup. Another is misapplying signs when solving for constants; always substitute the values that make each factor zero. Students also sometimes omit the absolute value inside logarithms. Remember ∫ 1/x dx = ln|x| + C, not ln x + C.
一个常见错误是遇到假分式时忘记先做多项式除法，这会导致部分分式的设立错误。另一个错误是在求解常数时弄错符号；始终代入能使各因式为零的数值。学生有时还会遗漏对数内的绝对值符号。记住 ∫ 1/x dx = ln|x| + C，而非 ln x + C。
For repeated linear factors, ensure your decomposition includes all descending powers from the exponent down to 1. Missing the lower power term will produce a system of equations with no solution or an incorrect result. Finally, practice switching between comparing coefficients and substitution—both are essential for exam success.
对于重复线性因式，要确保分解包含了从最高次幂一直降至一次幂的所有项。遗漏低次幂项将导致方程组无解或结果错误。最后，练习在对比系数法与代入法之间灵活切换——两者对考试成功都至关重要。
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