📚 Edexcel A-Level Integration: Tackling Question 138 (pdfjoiner 4) | Edexcel A-Level 积分：攻克第138题（pdfjoiner 4）
In this detailed revision walkthrough, we dissect a classic rational function integration problem that appears as Question 138 in a compiled Edexcel A-Level Mathematics practice set (pdfjoiner 4). The question requires a strategic combination of completing the square, numerator splitting, and recognising standard arctangent and logarithmic integrals – exactly the skills tested in the Pure Mathematics Paper 1 and Paper 2. Follow each section to master the technique and avoid the pitfalls that cost marks in the exam.
在此详细复习解析中，我们将剖析一道经典的有理函数积分题，它出现在一份Edexcel A-Level 数学综合练习集（pdfjoiner 4）的第138题。解答此题需要灵活搭配配方法、分子拆分以及识别标准的反正切和对数积分——这正是纯数试卷1和试卷2所考查的核心技能。跟随每一小节的讲解，掌握技巧并避开考试中导致失分的常见陷阱。
1. Problem Statement | 题目陈述
The task is to evaluate the indefinite integral ∫ (2x + 3) / (x² + 2x + 5) dx. This is a typical Edexcel question where the integrand is a rational function with a quadratic denominator that does not factorise over the real numbers, forcing you to use completing the square.
题目要求计算不定积分 ∫ (2x + 3)/(x² + 2x + 5) dx。这是一个典型的 Edexcel 考题，被积函数是有理函数，分母为二次式且无法在实数范围内因式分解，这迫使你必须使用配方法。
2. Identifying the Integral Type | 识别积分类型
would require real factorisation, which is impossible here because the discriminant Δ = 2² – 4 × 1 × 5 = 4 – 20 = – ent term. Recognising this pattern early saves time and guides the algebraic manipulation.
分子是一次式（次数1），分母是二次式（次数2），且没有公因式。直接部分分式需要实根因式分解，但此时判别式 Δ = 2² – 4 × 1 × 5 = 4 – 20 = –16 为负，故不可行。因此积分结果将包含自然对数项与反正切项的组合。尽早识别这一模式能节省时间，并指导代数变形方向。
3. Completing the Square for the Denominator | 对分母进行配方法
square: x² + 2x + 5 = (x² + 2x + 1) + 4 = (x + 1)² + 4. This transforms the integral into ∫ (2x + 3) / ((x + 1)² + 4) dx. The constant 4 can be written as 2², which is the standard form for an arctangent integral when the numerator is a constant, but here the numerator is still 2x+3.
通过配方改写分母：x² + 2x + 5 = (x² + 2x + 1) + 4 = (x + 1)² + 4。这样就把积分变形成 ∫ (2x + 3)/((x + 1)² + 4) dx。常数4可写成 2²，这正是反正切积分的标准形式（当分子为常数时），不过这里分子仍是 2x+3。
4. Splitting the Numerator Using the Derivative of the Denominator | 利用分母导数拆分分子
he denominator becomes u² + 4, and its derivative is 2u. We want the numerator 2x + 3 to be written as A·(derivative of u²+4) + B. Observe that d/du (u²+4) = 2u, and with u = x+1 we have 2u = 2(x+1). So we write 2x + 3 = 2(x+1) + 1 = 2u + 1. That is, the numerator is the derivative of the inner squared part plus an extra constant 1.
核心技巧是将分子用分母配方式内部变量的导数来表达。令 u = x + 1，则分母变成 u² + 4，其导数为 2u。我们希望将分子 2x + 3 写成 A·(u²+4的导数) + B 的形式。注意到 d/du (u²+4) = 2u，代入 u = x+1 得 2(x+1)。因此可写 2x + 3 = 2(x+1) + 1 = 2u + 1。也就是说，分子等于内层平方部分的导数再加一个常数1。
5. Separating the Integral into Two Parts | 将积分拆分为两部分
The integral now becomes ∫ [2(x+1) / ((x+1)² + 4)] dx + ∫ [1 / ((x+1)² + 4)] dx. The first integral has the form ∫ (f' 1)²+4 is 2(x+1). The second integral is a standard arctangent form after a simple scaling.
现在积分变为 ∫ [2(x+1)/((x+1)² + 4)] dx + ∫ [1/((x+1)² + 4)] dx。第一个积分形如 ∫ (f'(x)/f(x)) dx，因为分母 (x+1)²+4 的导数就是 2(x+1)。第二个积分经简单缩放后就是标准的反正切积分形式。
6. Evaluating the First Integral – Natural Logarithm | 计算第一个积分——自然对数
e the substitution w = (x+1)² + 4, so dw = 2(x+1) dx. The integral becomes ∫ (1/w) dw = ln|w| + C₁ = ln|(x+1)² + 4| ₁. Since (x+1)² + 4 is always positive, the absolute value can be omitted: ln((x+1)² + 4).
对 ∫ 2(x+1)/((x+1)² + 4) dx，令 w = (x+1)² + 4，则 dw = 2(x+1) dx。积分化为 ∫ (1/w) dw = ln|w| + C₁ = ln|(x+1)² + 4| + C₁。由于 (x+1)² + 4 恒为正，绝对值符号可去掉，写作 ln((x+1)² + 4)。
7. Evaluating the Second Integral – Arctangent | 计算第二个积分——反正切
dx requires the standard result ∫ 1/(u² + a²) du = (1/a) arctan(u/a). Here u = x+1, a² = 4 ⇒ 1/2) arctan((x+1)/2) + C₂. No extra substitution is needed beyond recognising the pattern.
积分 ∫ 1/((x+1)² + 4) dx 需要使用标准公式 ∫ 1/(u² + a²) du = (1/a) arctan(u/a)。这里 u = x+1，a² = 4 得 a = 2。因此 ∫ 1/((x+1)² + 2²) dx = (1/2) arctan((x+1)/2) + C₂。除识别模式外无需额外的代换。
8. Final Answer and Simplification | 最终答案与化简
Combining the two results and absorbing the constants into a single constant C gives the final answer:
∫ (2x + 3)/(x² + 2x + 5) dx = ln(x² + 2x + 5) + (1/2) arctan((x+1)/2) + C.
Notice that (x+1)² + 4 is exactly x² + 2x + 5, so the logarithm can be written directly in terms of the original denominator. Many examiners expect this tidy final form.
合并两个结果并将常数合并为一个常数 C，得到最终答案：
∫ (2x + 3)/(x² + 2x + 5) dx = ln(x² + 2x + 5) + (1/2) arctan((x+1)/2) + C。
注意 (x+1)² + 4 正是 x² + 2x + 5，因此对数项可直接写成原分母的自然对数。阅卷人通常期待这种整洁的最终形式。
9. Common Pitfalls and How to Avoid Them | 常见陷阱及规避方法
One frequent mistake is forgetting to split the numerator correctly – e arctangent: using a = 4 instead of a = 2, which gives (1/4) arctan((x+1)/4) and costs the marks. Also, dropping the dx when substituting, or writing ln|(x+1)²+4| th a superfluous absolute value, are minor errors that can be avoided by careful practice.
一个常见错误是分子拆分不当——学生往往试图将整个分子凑成导数，从而遗漏常数项。另一个错误是反正切部分的系数处理不当：误把 a = 4 当作 a = 2，得出 (1/4) arctan((x+1)/4) 而失分。此外，代换时漏掉 dx，或者在恒正表达式上画出多余的绝对值符号，都是通过细致练习可以避免的小错。
10. Practice Variation – Strengthen Your Fluency | 变式练习——提升熟练度
- 1)/(x² + 2x + 10) dx. Follow the same logic: complete the square to (x+1)²+9, split 4x-1 as 4(x+1) - 5, then integrate to obtain 2 ln(x²+2x+10) - (5/3) arctan((x+1)/3) + C. Creating your own variations and predicting the coefficients helps internalise the method.
为巩固技巧，尝试这道变式题：计算 ∫ (4x - 1)/(x² + 2x + 10) dx。遵循相同逻辑：配方得 (x+1)²+9，将 4x-1 拆分为 4(x+1) - 5，然后积分得到 2 ln(x²+2x+10) - (5/3) arctan((x+1)/3) + C。自己设计变体并预判系数有助于内化该方法。
11. Connection to Edexcel Mark Schemes | 与 Edexcel 评分标准的关联
In the Edexcel A-Level marking guidance, examiners award method marks (M1) for correctly completing the square and splitting the numerator, accuracy marks (A1) for each integrated term, and a final A1 for the simplified answer with +C. Even if an arithmetic slip occurs, structured working ensures you gain the majority of available marks.
在 Edexcel A-Level 的评分指导中，考官对正确配方和分子拆分给予方法分 (M1)，对每一项积分结果给予准确度分 (A1)，对带有 +C 的简化最终答案再给一个 A1。即使发生计算小失误，结构清晰的步骤也能确保你获得大部分分数。
12. Summary and Revision Takeaway | 总结与复习要点
rt is straightforward: complete the square → write the numerator as A·(derivative of squared term) + B → split into ln and arctan integrals → apply standard formulas → simplify. Having this routine committed to memory will make such problems a reliable source of marks on exam day.
第138题集中体现了 Edexcel 必考技能：对不可约二次分母的有理函数进行积分。解题流程非常直接：配方 → 将分子写成 A·(平方项导数) + B → 拆分为对数和反正切积分 → 套用标准公式 → 化简。牢记这一套流程，就能使这类题目成为考试中稳定的得分源。
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