📚 Solving Integration with Partial Fractions: Edexcel A-Level Problem (pdfjoiner_(4)-049) | 部分分式法解积分：Edexcel A-Level 习题 (pdfjoiner_(4)-049)
4)-049. The question requires a combination of algebraic manipulation and integration techniques, specifically partial fractions and logarithmic integration. Mastering such problems is essential for top marks in the Edexcel examination. We will walk through each step, highlight common pitfalls, and provide bilingual explanations to reinforce your understanding.
本文我们将攻克一道来自 Edexcel A-Level 纯数复习题集（编号 pdfjoiner_(4)-049）的高难度积分题。这道题要求综合运用代数变形与积分技巧，特别是部分分式与对数积分。掌握此类问题是 Edexcel 考试取得高分的关键。我们将逐步讲解每一步骤，指出常见错误，并提供双语解释以加深理解。
1. The Problem Statement | 题目陈述
Find the indefinite integral:
求不定积分：
∫ (2x + 1) / (x³ − x²) dx
This expression appears in many Edexcel past papers and is typical of the integration-by-partial-fractions category. The denominator x³ − x² can be factorised, allowing the fraction to be split into simpler terms. We must handle the repeated linear factor carefully.
该表达式常见于 Edexcel 历年真题，是部分分式积分类型的典型题目。分母 x³ − x² 可因式分解，从而使分式拆分为更简单的项。我们必须谨慎处理重复线性因子。
2. Factorising the Denominator | 分母因式分解
²(x − 1). The denominator now reveals a repeated linear factor x (with multiplicity 2) and a distinct linear factor (x − 1). This dictates the form of our partial fraction decomposition.
首先，提取分母中的最大公因子：x³ − x² = x²(x − 1)。此时分母展现出一个重复线性因子 x（重数为2）和一个相异的线性因子 (x − 1)。这决定了我们部分分式分解的形式。
3. Setting Up Partial Fractions | 建立部分分式
For a denominator containing x² and (x − 1), the correct partial fraction form is:
对于含有 x² 和 (x − 1) 的分母，正确的部分分式形式为：
(2x + 1) / [x²(x − 1)] = A/x + B/x² + C/(x − 1)
Here A, B, and C are constants to be determined. Note that a term with denominator x is included because of the repeated root; omitting it is a common mistake.
此处 A、B 和 C 为待定常数。请注意，由于存在重根，必须包含分母为 x 的项；漏写此项是常见错误。
4. Clearing Denominators | 去分母
Multiply both sides of the equation by the original denominator x²(x − 1) to obtain:
将方程两边同乘以原分母 x²(x − 1)，得到：
2x + 1 = A x(x − 1) + B (x − 1) + C x²
This polynomial identity must hold for all x. We will expand and compare coefficients, or substitute strategic values of x to find A, B, and C.
这一多项式恒等式对所有 x 均成立。我们可以展开并比较系数，或代入特定的 x 值来求出 A、B 和 C。
5. Finding the Constants | 求待定常数
Expand the right-hand side carefully:
仔细展开右侧：
A x(x − 1) = A x² − A x
B (x − 1) = B x − B
C x² = C x²
Summing these gives: (A + C)x² + (−A + B)x + (−B). Equating coefficients with 2x + 1 (which is 0·x² + 2x + 1) yields the system:
求和得：(A + C)x² + (−A + B)x + (−B)。与 2x + 1（即 0·x² + 2x + 1）比较系数，得到方程组：
- A + C = 0
- −A + B = 2
- −B = 1
Solving: From −B = 1, B = −1. Then −A + (−1) = 2 ⇒ −A = 3 ⇒ A = −3. Finally, A + C = 0 ⇒ −3 + C = 0 ⇒ C = 3.
求解：由 −B = 1 得 B = −1。代入 −A + (−1) = 2 ⇒ −A = 3 ⇒ A = −3。最后由 A + C = 0 ⇒ −3 + C = 0 ⇒ C = 3。
6. Rewriting the Integral | 重写积分式
Substitute the found constants back into the partial fractions:
将求得的常数代回部分分式：
(2x + 1) / [x²(x − 1)] = −3/x − 1/x² + 3/(x − 1)
Thus the original integral becomes a sum of three simpler integrals:
因此原积分化为三个更简单积分的和：
∫ (−3/x) dx + ∫ (−1/x²) dx + ∫ 3/(x − 1) dx
Each of these can be integrated directly using standard results.
这些积分都可以直接使用标准结果进行积分。
7. Integrating Each Term | 逐项积分
Recall basic integration rules:
回顾基本积分法则：
- ∫ (1/x) dx = ln |x| + C
- ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, for n ≠ −1
Apply these:
应用如下：
First term: ∫ (−3/x) dx = −3 ln |x|
第一项： ∫ (−3/x) dx = −3 ln |x|
Second term: ∫ (−1/x²) dx = ∫ (−x⁻²) dx = −(x⁻¹/(−1)) = 1/x (since −x⁻² integrates to x⁻¹)
第二项： ∫ (−1/x²) dx = ∫ (−x⁻²) dx = −(x⁻¹/(−1)) = 1/x （因为 −x⁻² 的积分是 x⁻¹）
More precisely: ∫ −x⁻² dx = ∫ −x⁻² dx = −(x⁻¹/−1) = x⁻¹ = 1/x.
更准确地：∫ −x⁻² dx = −(x⁻¹/−1) = x⁻¹ = 1/x。
Third term: ∫ 3/(x − 1) dx = 3 ln |x − 1|
第三项： ∫ 3/(x − 1) dx = 3 ln |x − 1|
Don't forget the constant of integration + C at the end.
最后不要忘记积分常数 + C。
8. Assembling the Final Answer | 组合最终答案
Combine the three results:
将三项结果组合：
∫ (2x + 1)/(x³ − x²) dx = −3 ln |x| + 1/x + 3 ln |x − 1| + C
This can be tidied using logarithm rules, though it is not required. Sometimes the answer is expressed as:
可以使用对数法则进行整理，但这并非必需。有时答案表示为：
1/x + 3 ln | (x − 1)/x | + C
Both forms are acceptable in Edexcel mark schemes.
两种形式在 Edexcel 评分标准中均可接受。
9. Verification by Differentiation | 通过求导验证
To check the correctness, differentiate the antiderivative:
为检验正确性，对原函数求导：
d/dx [−3 ln |x| + 1/x + 3 ln |x − 1|] = −3/x − 1/x² + 3/(x − 1).
This sum simplifies back to (2x + 1)/[x²(x − 1)], confirming our result. Always perform this quick check if time allows in an exam.
该和式可化简回 (2x + 1)/[x²(x − 1)]，证实了我们的结果。考试中若时间允许，务必进行此项快速检验。
10. Common Pitfalls | 常见易错点
- Forgetting the A/x term when handling repeated factors like x². The correct form must include denominators of all descending powers from the repeated root.
- Mishandling signs when expanding and equating coefficients; double-check your algebra.
- tegration of 1/x²: remember ∫ x⁻² dx = −x⁻¹, so the negative sign in front flips the sign.
- Missing the absolute values inside logarithms; Edexcel expects ln|...| for indefinite integrals.
- Omitting the constant of integration + C, which costs a mark.
- 处理 x² 这样的重因子时忘记 A/x 项。正确的形式必须包含自重根开始的全部递减幂次的分母。
- 展开和比较系数时符号处理错误；要仔细核对代数步骤。
- 对 1/x² 的积分出错：记住 ∫ x⁻² dx = −x⁻¹，因此前面的负号会改变正负。
- 对数内遗漏绝对值；Edexcel 期望不定积分中使用 ln|...|。
- 遗漏积分常数 + C，导致失分。
11. Exam Tips for Similar Questions | 类似题型的应试技巧
In the exam, you might be asked to evaluate a definite integral after partial fractions, or to first perform long division if the degree of the numerator is equal to or greater than the denominator. Look carefully at the algebraic structure and factorise completely. If a substitution is hinted, follow it, but with this type of problem, direct partial fractions is swift and reliable.
考试中，可能要求在部分分式后计算定积分，或者当分子次数不低于分母时先做多项式长除。仔细观察代数结构并完全因式分解。若题目提示换元，则按提示操作，但对于此类问题，直接使用部分分式法快速且可靠。
12. Practice Variation | 变式练习
) dx. Remember to factorise the quadratic denominator, set up partial fractions, and integrate. The method mirrors what we have done, but with distinct linear factors, which is even more straightforward.
尝试以下类似问题来巩固技巧：求 ∫ (3x − 1) / (x² + x − 2) dx。记得对二次分母因式分解，建立部分分式并积分。方法与我们刚才的操作相仿，但由于是不同的线性因子，反而更简单。
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