📚 pdfjoiner_(4)-101: Partial Fractions with a Repeated Linear Factor | 含重线性因式的部分分式积分
In this article, we take a close look at a problem from the pdfjoiner practice set — question 101 from section 4. The task is to evaluate an indefinite integral of a rational function whose denominator contains a repeated linear factor. Mastering this type of integral is essential for success in Edexcel A-Level Mathematics, as it brings together algebraic manipulation, partial fractions and elementary integration techniques. We will work through every step, from factorising the denominator to verifying the final result, ensuring you gain a deep understanding of the method.
在这篇文章中，我们将仔细分析 pdfjoiner 练习集中的一道题目——第4部分第101题。任务是求解一个分母含有重线性因式的有理函数的不定积分。掌握这类积分对 Edexcel A-Level 数学考试至关重要，因为它综合了代数变形、部分分式分解和基本积分技巧。我们将逐步讲解，从分母因式分解到验证最终结果，确保你彻底理解这一方法。
1. Problem Statement | 问题陈述
The problem asks us to find the indefinite integral: ∫ (x² + 3x + 5) / ((x – minator (3). The denominator is already partially factorised, but we must recognise that the factor (x – g it a repeated linear factor. This repetition requires a specific partial fraction setup.
题目要求计算不定积分：∫ (x² + 3x + 5) / ((x – 1)² (x + 2)) dx。乍看之下，被积函数是一个真分式，因为分子的次数（2）低于分母的次数（3）。分母已经部分因式分解，但我们必须注意到 (x – 1) 出现了两次，是一个重线性因式。这种重复需要设置特殊形式的部分分式。
2. Recognising the Denominator Structure | 识别分母结构
Let's write the denominator explicitly: (x – 1)²(x + 2). It has two distinct roots: x = 1 (multiplicity 2) and x = –2 (multiplicity 1). In partial fractions, a repeated factor of the form (x – a)ⁿ contributes a sum of terms A₁/(x – a) + A₂/(x – a)² + ... + Aₙ/(x – a)ⁿ. Since our exponent is 2, we need two terms for the (x – 1) factor: one with denominator (x – 1) and one with denominator (x – 1)². The distinct factor (x + 2) gets a single term.
我们将分母明确写出：(x – 1)²(x + 2)。它有两个不同的根：x = 1（重数为2）和 x = –2（重数为1）。在部分分式中，形如 (x – a)ⁿ 的重因式会产生一组项：A₁/(x – a) + A₂/(x – a)² + … + Aₙ/(x – a)ⁿ。由于这里的指数是2，我们需要为 (x – 1) 设置两项：一项分母为 (x – 1)，另一项分母为 (x – 1)²。而单独的因式 (x + 2) 只对应一项。
3. General Form of Partial Fractions | 部分分式的一般形式
We propose the decomposition: (x² + 3x + 5) / ((x – 1)²(x + 2)) = A/(x – 1) + B/(x – nominator. This setup covers all possible proper fractions with this denominator. The next task is to find the numerical values of A, B and C.
我们设分解形式为：(x² + 3x + 5) / ((x – 1)²(x + 2)) = A/(x – 1) + B/(x – 1)² + C/(x + 2)，其中 A、B、C 为待定常数。注意 B 是平方项的分子的系数，而 A 对应一次因式的分母。这种设定涵盖了具有该分母的所有可能真分式。接下来的任务是求出 A、B、C 的具体数值。
4. Equating Numerators | 使分子相等
Multiply both sides by the common denominator (x – ide becomes simply the original numerator x² + 3x + 5. The right-hand side expands as: A(x – 1)(x + 2) + B(x + 2) + C(x – 1)². Let us expand each term carefully: A(x² + x – 2) = Ax² + Ax – 2A; B(x + 2) = Bx + 2B; C(x² – 2x + 1) = Cx² – 2Cx + C. Summing them yields (A + C)x² + (A + B – 2C)x + (– + C). This polynomial must be identically equal to x² + 3x + 5, so we equate coefficients.
两边同时乘上公分母 (x – 1)²(x + 2) 以去掉分数。左边就是原来的分子 x² + 3x + 5。右边展开为：A(x – 1)(x + 2) + B(x + 2) + C(x – 1)²。我们仔细展开每一项：A(x² + x – 2) = Ax² + Ax – 2A；B(x + 2) = Bx + 2B；C(x² – 2x + 1) = Cx² – 2Cx + C。将它们相加得到 (A + C)x² + (A + B – 2C)x + (–2A + 2B + C)。这个多项式必须恒等于 x² + 3x + 5，因此我们比较系数。
5. Solving for Constants | 求解系数
Comparing coefficients of x², x and the constant term gives a system of three equations:
比较 x² 项、x 项和常数项的系数，得到如下三元一次方程组：
| A + C = 1 | (coefficient of x²) |
| A + B – 2C = 3 | (coefficient of x) |
| –2A + 2B + C = 5 | (constant term) |
From the first equation, C = 1 – A. Substitute into the second: A + B – 2(1 – A) = 3 → A + B – 2 + 2A = 3 → 3A + B = 5. Substitute into the third: –2A + 2B + (1 – A) = 5 → –3A + 2B = 4. Now solve the two equations 3A + B = 5 and –3A + 2B = 4 simultaneously. Adding them eliminates A: (3A – 2B) = 5 + 4 → 3B = 9 → B = 3. Then from 3A + 3 = 5 we get 3A = 2 → A = 2/3. Finally, C = 1 – 2/3 = 1/3. The constants are A = 2/3, B = 3, C = 1/3.
由第一个方程得 C = 1 – A。代入第二个方程：A + B – 2(1 – A) = 3 → A + B – 2 + 2A = 3 → 3A + B = 5。代入第三个方程：–2A + 2B + (1 – A) = 5 → –3A + 2B = 4。现在联立求解 3A + B = 5 和 –3A + 2B = 4。两式相加消去 A：(3A – 3A) + (B + 2B) = 5 + 4 → 3B = 9 → B = 3。然后由 3A + 3 = 5 得 3A = 2 → A = 2/3。最后 C = 1 – 2/3 = 1/3。所求常数为 A = 2/3，B = 3，C = 1/3。
6. Writing the Decomposed Integral | 写出分解后的积分
With the constants found, the partial fraction decomposition becomes: (x² + 3x + 5)/((x – 1)²(x + 2)) = (2/3)/(x – 1) + 3/(x – . Therefore, the original integral splits into three separate simpler integrals: ∫ (2/3)(x – 1)⁻¹ dx + ∫ 3(x – his is much easier to handle because each term corresponds to a standard integration rule.
求出常数后，部分分式分解变为：(x² + 3x + 5)/((x – 1)²(x + 2)) = (2/3)/(x – 1) + 3/(x – 1)² + (1/3)/(x + 2)。于是原积分拆分为三个更简单的积分：∫ (2/3)(x – 1)⁻¹ dx + ∫ 3(x – 1)⁻² dx + ∫ (1/3)(x + 2)⁻¹ dx。这样处理起来容易得多，因为每一项都符合一个标准的积分公式。
7. Integrating Simple Fractions | 积分简单分式
Recall two basic integration results: ∫ (1/u) du = ln|u| + C and ∫ u⁻ⁿ du = u¹⁻ⁿ/(1 – we have a linear substitution. For our terms, the first integral is of the form ∫ (2/3)(x – 1)⁻¹ dx. Let u = x – 1, then du = dx, giving (2/3) ln|u| = (2/3) ln|x – 1|. The third integral is similar: ∫ (1/3)(x + 2)⁻¹ dx = (1/3) ln|x + 2|. The second term requires the power rule: ∫ 3(x – 1)⁻² dx = 3 ∫ u⁻² du = 3(u⁻¹/(–1)) = –3/u = –3/(x – 1).
回顾两个基本积分结果：∫ (1/u) du = ln|u| + C，以及当 n ≠ 1 时，∫ u⁻ⁿ du = u¹⁻ⁿ/(1 – n) + C。对于第一类积分，我们使用线性换元。在我们的式子中，第一个积分形如 ∫ (2/3)(x – 1)⁻¹ dx。令 u = x – 1，则 du = dx，得到 (2/3) ln|u| = (2/3) ln|x – 1|。第三个积分类似：∫ (1/3)(x + 2)⁻¹ dx = (1/3) ln|x + 2|。第二项则需使用幂法则：∫ 3(x – 1)⁻² dx = 3 ∫ u⁻² du = 3(u⁻¹/(–1)) = –3/u = –3/(x – 1)。
8. Integrating the (x – 1)⁻² Term | 积分 (x – 1)⁻² 项
r often causes sign errors, let’s isolate the second term once more. The integral is ∫ 3(x – 1)⁻² dx. Factor out the constant 3: 3 ∫ (x – 1)⁻² dx. Substitute u = x – 1, du = dx. The integral of u⁻² is –u⁻¹, so the result is 3(–u⁻¹) = –3u⁻¹ = –3/(x – 1). Do not forget the minus sign — it arises from the formula ∫ u⁻² du = –1/u. This step is a common source of lost marks in exams, so double-check your work.
由于负次幂的积分常导致符号错误，我们再次单独处理第二项。积分为 ∫ 3(x – 1)⁻² dx。将常数 3 提到外面：3 ∫ (x – 1)⁻² dx。令 u = x – 1，du = dx。u⁻² 的积分是 –u⁻¹，于是结果为 3(–u⁻¹) = –3u⁻¹ = –3/(x – 1)。不要漏掉负号——它来自公式 ∫ u⁻² du = –1/u。这一步是考试中常见的丢分点，务必仔细检查。
9. Integrating Logarithmic Terms | 积分对数项
Now handle the two logarithmic integrals together. The first gives (2/3) ln|x – 1|, and the third gives (1/3) ln|x + 2|. Both absolute value bars are necessary because the natural logarithm is only defined for positive arguments; we use the modulus to ensure the antiderivative is valid on both sides of the asymptotes, provided we restrict to intervals that do not cross them. We can also combine the logs later if helpful, but for clarity we will keep them separate.
现在一并处理两个对数积分。第一个得到 (2/3) ln|x – 1|，第二个得到 (1/3) ln|x + 2|。两个绝对值符号都是必要的，因为自然对数只对正的自变量有定义；我们使用模长来保证原函数在渐近线两侧的区间上有效，前提是我们限制在跨越渐近线的区间上。如果方便，以后也可以将两个对数合并，但为了清晰起见，我们暂时分开书写。
10. Combining the Results | 合并结果
rts and including the constant of integration +C, the final answer is: ∫ (x² + 3x + 5)/((x – 1)²(x + 2)) dx = (2/3) ln|x – 1|
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