A-Level Integration Techniques — A-Level 积分技巧

📚 A-Level Integration Techniques | A-Level 积分技巧

Integration is one of the cornerstones of A-Level Mathematics. While differentiation follows a fairly mechanical set of rules, integration demands creativity, pattern recognition, and a deep understanding of function behaviour. In this article, we explore the three essential integration techniques that every A-Level student must master: integration by substitution, integration by parts, and integration using partial fractions. Each technique is introduced with worked examples, common pitfalls, and exam tips.

积分是 A-Level 数学的核心支柱之一。如果说微分遵循一套相对机械的规则，那么积分则需要创造力、模式识别和对函数行为的深刻理解。本文探讨 A-Level 学生必须掌握的三种核心积分技巧：换元积分法、分部积分法以及部分分式积分法。每种技巧都配有详细例题、常见陷阱和考试技巧。

1. Integration by Substitution | 换元积分法

1.1 The Core Idea | 核心思想

Integration by substitution is essentially the chain rule in reverse. When you see a composite function — a function inside another function — substitution can simplify the integral dramatically. The idea is to let u be the inner function, then express everything (including dx) in terms of u and du.

换元积分法本质上是链式法则的逆运算。当你看到一个复合函数——一个函数嵌套在另一个函数里——换元法可以极大地简化积分。核心思路是令 u 等于内层函数，然后将所有变量（包括 dx）用 u 和 du 表示。

The general formula is:

一般公式为：

∫ f(g(x)) · g'(x) dx = ∫ f(u) du, where u = g(x)

1.2 Worked Example 1 | 例题 1

Evaluate ∫ 2x(x² + 1)⁴ dx

计算 ∫ 2x(x² + 1)⁴ dx

Let u = x² + 1. Then du/dx = 2x, so du = 2x dx. The integral becomes ∫ u⁴ du = u⁵/5 + C. Substituting back: (x² + 1)⁵ / 5 + C.

令 u = x² + 1，则 du/dx = 2x，因此 du = 2x dx。积分变为 ∫ u⁴ du = u⁵/5 + C。代回 x：(x² + 1)⁵ / 5 + C。

1.3 Worked Example 2: Trigonometric Substitution | 例题 2：三角换元

Evaluate ∫ sin³x · cos x dx

计算 ∫ sin³x · cos x dx

Let u = sin x. Then du/dx = cos x, so du = cos x dx. The integral becomes ∫ u³ du = u⁴/4 + C = sin⁴x / 4 + C.

令 u = sin x，则 du/dx = cos x，因此 du = cos x dx。积分变为 ∫ u³ du = u⁴/4 + C = sin⁴x / 4 + C。

1.4 Definite Integrals and Changing Limits | 定积分与换限

For definite integrals, you have two options: (1) change the limits when you change the variable, or (2) substitute back to x before evaluating. Option (1) is generally faster and less error-prone.

对于定积分，你有两种选择：(1) 在换元时同时改变积分限，或 (2) 先代回 x 再代入原积分限。选项 (1) 通常更快且更不容易出错。

Example: ∫₀¹ 2x(x² + 1)⁴ dx

When x = 0, u = 0² + 1 = 1. When x = 1, u = 1² + 1 = 2. So the integral becomes ∫₁² u⁴ du = [u⁵/5]₁² = 32/5 − 1/5 = 31/5.

当 x = 0 时，u = 0² + 1 = 1。当 x = 1 时，u = 1² + 1 = 2。因此积分变为 ∫₁² u⁴ du = [u⁵/5]₁² = 32/5 − 1/5 = 31/5。

1.5 Common Pitfall: Forgetting du/dx | 常见陷阱：忘记 du/dx

The most frequent mistake is forgetting to convert dx into du. Students often write du instead of substituting properly. Always check: does your du match the extra factor outside the main function? If not, you may need a different u or a different technique entirely.

最常见的错误是忘记将 dx 转换为 du。学生经常写错 du 的表达式。切记检查：你的 du 是否与主要函数之外的额外因子匹配？如果不匹配，你可能需要选择不同的 u 或完全不同的方法。

2. Integration by Parts | 分部积分法

2.1 The Formula and Its Origin | 公式及其来源

Integration by parts comes from the product rule for differentiation. If d(uv)/dx = u · dv/dx + v · du/dx, then rearranging and integrating both sides gives:

分部积分法源于微分的乘法法则。如果 d(uv)/dx = u · dv/dx + v · du/dx，那么重排并积分两边得到：

∫ u dv = uv − ∫ v du

In practice, you choose one part of the integrand as u (which becomes simpler when differentiated) and the rest as dv (which you can integrate).

实际应用中，你选择被积函数的一部分作为 u（它在求导后变得更简单），其余部分作为 dv（你能够积分的部分）。

2.2 The LIATE Rule | LIATE 法则

How do you choose u? Use the LIATE priority order, from highest to lowest priority for u:

如何选择 u？使用 LIATE 优先级顺序，从高到低确定 u 的选择：

Priority	Function Type	Examples
L	Logarithmic	ln x, log₂ x
I	Inverse Trigonometric	arcsin x, arctan x
A	Algebraic (Polynomial)	x², x³ + 2x
T	Trigonometric	sin x, cos x, tan x
E	Exponential	eˣ, 2ˣ

L has the highest priority for u, E the lowest. So for ∫ x eˣ dx, u = x (algebraic, higher priority than exponential) and dv = eˣ dx.

L 的优先级最高（最适合做 u），E 的优先级最低。因此对于 ∫ x eˣ dx，u = x（代数函数，优先级高于指数函数），dv = eˣ dx。

2.3 Worked Example 1: Polynomial × Exponential | 例题 1：多项式 × 指数函数

Evaluate ∫ x eˣ dx

计算 ∫ x eˣ dx

By LIATE: u = x, dv = eˣ dx. Then du = dx, v = eˣ. Applying the formula: ∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C.

根据 LIATE：u = x，dv = eˣ dx。则 du = dx，v = eˣ。应用公式：∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C。

2.4 Worked Example 2: Logarithmic Function | 例题 2：对数函数

Evaluate ∫ ln x dx

计算 ∫ ln x dx

There appears to be only one function here, but we can treat it as ∫ 1 · ln x dx. By LIATE: u = ln x (L, highest priority), dv = dx. Then du = (1/x) dx, v = x. So ∫ ln x dx = x ln x − ∫ x · (1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C.

这里看似只有一个函数，但我们可以把它看作 ∫ 1 · ln x dx。根据 LIATE：u = ln x（L，最高优先级），dv = dx。则 du = (1/x) dx，v = x。因此 ∫ ln x dx = x ln x − ∫ x · (1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C。

2.5 Worked Example 3: Repeated Integration by Parts | 例题 3：重复分部积分

Evaluate ∫ x² sin x dx

计算 ∫ x² sin x dx

Sometimes one round is not enough. Here u = x², dv = sin x dx. Then du = 2x dx, v = −cos x. First round: ∫ x² sin x dx = −x² cos x + ∫ 2x cos x dx. Now apply integration by parts again to ∫ 2x cos x dx, with u = 2x, dv = cos x dx. Second round gives: ∫ 2x cos x dx = 2x sin x − ∫ 2 sin x dx = 2x sin x + 2 cos x. Final answer: −x² cos x + 2x sin x + 2 cos x + C.

有时一轮分部积分不够。这里 u = x²，dv = sin x dx。则 du = 2x dx，v = −cos x。第一轮：∫ x² sin x dx = −x² cos x + ∫ 2x cos x dx。现在对 ∫ 2x cos x dx 再次应用分部积分，令 u = 2x，dv = cos x dx。第二轮得到：∫ 2x cos x dx = 2x sin x − ∫ 2 sin x dx = 2x sin x + 2 cos x。最终答案：−x² cos x + 2x sin x + 2 cos x + C。

2.6 Common Pitfall: Choosing the Wrong u | 常见陷阱：选错 u

If you reverse the LIATE rule — for example, choosing u = eˣ and dv = x dx in ∫ x eˣ dx — the integral becomes more complicated: ∫ x eˣ dx = (x²/2)eˣ − ∫ (x²/2)eˣ dx. Now you have x² in the remaining integral, which is even harder. Always apply LIATE correctly.

如果你把 LIATE 法则用反了——例如在 ∫ x eˣ dx 中选择 u = eˣ 和 dv = x dx——积分会变得更复杂：∫ x eˣ dx = (x²/2)eˣ − ∫ (x²/2)eˣ dx。现在剩下的积分中出现了 x²，更难处理。一定要正确使用 LIATE 法则。

3. Integration Using Partial Fractions | 部分分式积分法

3.1 The Core Idea | 核心思想

Partial fractions decompose a rational function (one polynomial divided by another) into a sum of simpler fractions. Each simpler fraction can then be integrated using standard results — typically yielding logarithms or arctangents.

部分分式将有理函数（一个多项式除以另一个多项式）分解为多个简单分式的和。每个简单分式都可以用标准公式积分——通常会得到对数或反正切函数。

3.2 The Method Step-by-Step | 方法步骤

Step 1: Ensure the degree of the numerator is less than the degree of the denominator. If not, perform polynomial long division first.

步骤 1：确保分子的次数低于分母的次数。如果不是，先进行多项式长除。

Step 2: Factorise the denominator completely into linear factors (ax + b) and irreducible quadratic factors (ax² + bx + c).

步骤 2：将分母完全因式分解为线性因子 (ax + b) 和不可约的二次因子 (ax² + bx + c)。

Step 3: Write the partial fraction decomposition with unknown constants.

步骤 3：写出包含未知常数的部分分式分解形式。

Step 4: Solve for the constants by equating coefficients or substituting convenient values of x.

步骤 4：通过比较系数或代入方便的 x 值来求出这些常数。

Step 5: Integrate each term separately.

步骤 5：分别积分每一项。

3.3 Case 1: Distinct Linear Factors | 情况 1：不同的线性因子

Evaluate ∫ (2x + 1) / [(x − 1)(x + 2)] dx

计算 ∫ (2x + 1) / [(x − 1)(x + 2)] dx

Write: (2x + 1) / [(x − 1)(x + 2)] = A/(x − 1) + B/(x + 2). Multiply through by (x − 1)(x + 2): 2x + 1 = A(x + 2) + B(x − 1). Substitute x = 1: 3 = 3A ⇒ A = 1. Substitute x = −2: −3 = −3B ⇒ B = 1. So the integral becomes ∫ [1/(x − 1) + 1/(x + 2)] dx = ln|x − 1| + ln|x + 2| + C = ln|(x − 1)(x + 2)| + C.

设：(2x + 1) / [(x − 1)(x + 2)] = A/(x − 1) + B/(x + 2)。两边同乘 (x − 1)(x + 2)：2x + 1 = A(x + 2) + B(x − 1)。代入 x = 1：3 = 3A ⇒ A = 1。代入 x = −2：−3 = −3B ⇒ B = 1。因此积分变为 ∫ [1/(x − 1) + 1/(x + 2)] dx = ln|x − 1| + ln|x + 2| + C = ln|(x − 1)(x + 2)| + C。

3.4 Case 2: Repeated Linear Factors | 情况 2：重复的线性因子

For a factor like (x − a)ⁿ, you need terms with denominators (x − a), (x − a)², …, (x − a)ⁿ.

对于像 (x − a)ⁿ 这样的因子，你需要分母分别为 (x − a)、(x − a)²… (x − a)ⁿ 的各项。

Evaluate ∫ (3x + 5) / (x + 1)² dx

计算 ∫ (3x + 5) / (x + 1)² dx

Write: (3x + 5)/(x + 1)² = A/(x + 1) + B/(x + 1)². Multiply: 3x + 5 = A(x + 1) + B = Ax + (A + B). Equating coefficients: for x: A = 3; constant: A + B = 5 ⇒ 3 + B = 5 ⇒ B = 2. So the integral becomes ∫ [3/(x + 1) + 2/(x + 1)²] dx = 3 ln|x + 1| − 2/(x + 1) + C.

设：(3x + 5)/(x + 1)² = A/(x + 1) + B/(x + 1)²。两边同乘：3x + 5 = A(x + 1) + B = Ax + (A + B)。比较系数：x 项：A = 3；常数项：A + B = 5 ⇒ 3 + B = 5 ⇒ B = 2。因此积分变为 ∫ [3/(x + 1) + 2/(x + 1)²] dx = 3 ln|x + 1| − 2/(x + 1) + C。

3.5 Case 3: Irreducible Quadratic Factor | 情况 3：不可约二次因子

When the denominator has an irreducible quadratic factor ax² + bx + c (where b² − 4ac < 0), the numerator must be linear: (Ax + B)/(ax² + bx + c). Integrating this typically involves completing the square and using the arctangent formula.

当分母包含不可约二次因子 ax² + bx + c（其中 b² − 4ac < 0）时，分子必须是一次式：(Ax + B)/(ax² + bx + c)。积分这类式子通常需要配方并使用反正切公式。

Evaluate ∫ (2x + 3) / (x² + 4x + 13) dx

计算 ∫ (2x + 3) / (x² + 4x + 13) dx

Complete the square: x² + 4x + 13 = (x + 2)² + 9. Now write numerator to match: 2x + 3 = 2(x + 2) − 1. The integral splits into ∫ 2(x + 2)/[(x + 2)² + 9] dx − ∫ 1/[(x + 2)² + 9] dx. First part: let u = (x + 2)² + 9, du = 2(x + 2) dx ⇒ ln|(x + 2)² + 9|. Second part: (1/3) arctan[(x + 2)/3]. Final: ln|x² + 4x + 13| − (1/3) arctan[(x + 2)/3] + C.

配方：x² + 4x + 13 = (x + 2)² + 9。将分子重写以匹配：2x + 3 = 2(x + 2) − 1。积分拆分为 ∫ 2(x + 2)/[(x + 2)² + 9] dx − ∫ 1/[(x + 2)² + 9] dx。第一部分：令 u = (x + 2)² + 9，du = 2(x + 2) dx ⇒ ln|(x + 2)² + 9|。第二部分：(1/3) arctan[(x + 2)/3]。最终结果：ln|x² + 4x + 13| − (1/3) arctan[(x + 2)/3] + C。

4. Choosing the Right Technique | 选择正确的技巧

Students often struggle most not with executing the techniques, but with knowing which one to apply. Here is a decision flowchart distilled from hundreds of A-Level exam problems:

学生最常遇到的困难往往不是执行技巧本身，而是不知道该用哪一种。以下是从数百道 A-Level 真题中提炼出的决策流程图：

Pattern in Integrand	Recommended Technique	Example
f(g(x)) · g'(x) — clear inner derivative	Substitution	∫ 2x sin(x²) dx
Product of two different function types	Integration by Parts	∫ x ln x dx, ∫ x eˣ dx
Rational function (polynomial/polynomial)	Partial Fractions	∫ (x+1)/(x²−1) dx
Single ln or inverse trig	Integration by Parts (dv = dx)	∫ ln x dx
Trigonometric powers	Substitution (often)	∫ sin³x cos x dx
eᵏˣ × sin/cos	Parts (twice, then solve)	∫ eˣ sin x dx

5. Exam Strategy and Common Mistakes | 考试策略与常见错误

5.1 The Constant of Integration | 积分常数

Never forget +C for indefinite integrals. In A-Level exams, omitting the constant can cost you a mark even if the rest is perfect. For definite integrals, make sure you evaluate both limits correctly — this is where most arithmetic mistakes occur.

千万不要忘记不定积分的 +C。在 A-Level 考试中，即使其他部分完全正确，遗漏常数项也可能丢分。对于定积分，确保正确代入两个积分限——这是算术错误最常出现的地方。

5.2 Sign Errors in Integration by Parts | 分部积分中的符号错误

The formula is ∫ u dv = uv − ∫ v du. The minus sign is critical. A common slip is writing + instead of −, especially after multiple rounds of integration by parts. Double-check your signs at every step.

公式是 ∫ u dv = uv − ∫ v du。减号至关重要。一个常见的失误是在多轮分部积分后把减号写成加号。每一步都要仔细检查符号。

5.3 Checking Your Answer | 检验你的答案

The beauty of integration is that you can always verify your result by differentiating it. If the derivative of your answer gives back the original integrand, you are correct. This is the single most powerful exam technique — use it whenever you have time.

积分的美妙之处在于你总是可以通过求导来验证结果。如果你答案的导数等于原来的被积函数，你就是对的。这是最强大的考试技巧——只要时间允许，一定要用它来检验。

5.4 The “Special Case” Trap | “特殊情况”陷阱

Some integrals look like they need one technique but actually require another. For example, ∫ 1/(x ln x) dx looks like a candidate for partial fractions, but it is actually solved by the substitution u = ln x, giving ∫ 1/u du = ln|ln x| + C. Always scan the integral for a f'(x)/f(x) pattern before diving into more complex methods.

有些积分表面上看起来需要某种技巧，但实际上需要另一种。例如，∫ 1/(x ln x) dx 看起来像是部分分式的候选者，但实际上用换元 u = ln x 即可解决，得到 ∫ 1/u du = ln|ln x| + C。在深入使用更复杂的方法之前，始终先检查积分中是否存在 f'(x)/f(x) 的模式。

6. Summary | 总结

Mastering integration techniques is a journey of pattern recognition. The three pillars — substitution, integration by parts, and partial fractions — cover the vast majority of A-Level integration problems. Practice is essential: start with textbook exercises that tell you which method to use, then progress to mixed exercises where you must decide the approach yourself. Remember, the best mathematicians are not those who never make mistakes, but those who catch their mistakes by verifying their answers.

掌握积分技巧是一个模式识别的过程。三大支柱——换元法、分部积分法和部分分式法——涵盖了 A-Level 中绝大部分积分问题。练习至关重要：从标注了使用方法的课本习题开始，然后过渡到需要你自己决定方法的混合习题。请记住，最好的数学家不是从不犯错的人，而是那些通过验证答案来发现并纠正错误的人。

Key takeaway: Before you start integrating, pause and look at the integrand. Is there a composite function with its derivative sitting next to it? (Substitution.) Is it a product of two distinct function types? (Parts.) Is it a rational function? (Partial fractions.) This three-second scan will save you minutes of wasted work and countless marks in your exam.

核心要点：在开始积分之前，停下来审视被积函数。是否有复合函数与其导数紧挨着？（换元法。）是否是两种不同函数类型的乘积？（分部积分法。）是否是有理函数？（部分分式法。）这三秒钟的扫描将为你节省数分钟的无用功，并在考试中避免大量失分。