Integration Techniques for Edexcel A-Level Pure 4 | Edexcel A-Level Pure 4 积分技巧

📚 Integration Techniques for Edexcel A-Level Pure 4 | Edexcel A-Level Pure 4 积分技巧

Integration techniques form the backbone of Edexcel A-Level Pure Mathematics 4. In this topic, you learn to handle integrals that cannot be evaluated using standard results alone. You will need to combine substitution, integration by parts and partial fractions to succeed in the exam.

积分技巧是 Edexcel A-Level Pure Mathematics 4 的核心内容。在本主题中，你将学习如何处理仅靠标准公式无法求解的积分。想在考试中取得好成绩，必须综合运用换元法、分部积分法和部分分式法。

1. Integration as the Reverse of Differentiation | 积分是微分的逆运算

Integration is the reverse process of differentiation, but unlike differentiation, there is no single rule that works for every function. You need a toolbox of methods to deal with products, quotients, composite functions and rational expressions.

积分是微分的逆运算，但与微分不同，没有一条规则适用于所有函数。你需要一套方法工具箱来处理乘积、商、复合函数和有理式。

dy/dx = f(x) ⇒ y = ∫ f(x) dx + C

Always remember the constant of integration when evaluating an indefinite integral. In Edexcel exams, omitting +C can lose a mark, especially when a general solution is required.

在计算不定积分时，务必记住积分常数。在 Edexcel 考试中，遗漏 +C 可能丢分，尤其是在需要求通解的情况下。

2. Integration by Substitution | 换元积分法

Substitution is used when the integrand contains a composite function and the derivative of the inner function is present, or can be arranged. You replace the inner function by u and rewrite dx in terms of du.

当被积函数包含复合函数且内层函数的导数存在（或可以凑出）时，使用换元法。将内层函数设为 u，并用 du 重写 dx。

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

For example, to evaluate ∫ 2x·sin(x²) dx, let u = x². Then du/dx = 2x, so du = 2x dx. The integral becomes ∫ sin u du = -cos u + C = -cos(x²) + C.

例如，计算 ∫ 2x·sin(x²) dx 时，令 u = x²。则 du/dx = 2x，所以 du = 2x dx。原积分变为 ∫ sin u du = -cos u + C = -cos(x²) + C。

In many Edexcel questions, the substitution is given. You must show every step: state u, find du/dx, replace dx, rewrite the integral and then integrate in terms of u before substituting back.

在 Edexcel 的许多题目中，会给出换元式。你必须展示每一步：写出 u，求出 du/dx，替换 dx，用 u 重写积分，然后在代回原变量前对 u 积分。

3. Definite Integrals by Substitution | 定积分的换元法

For definite integrals, you can either change the limits to u-values before evaluating, or substitute back to x before applying the original limits. Changing limits usually saves time and reduces sign errors.

对于定积分，可以先将上下限转换为 u 值再计算，也可以代回 x 后再代原上下限。转换上下限通常更省时且减少符号错误。

∫ from x = a to x = b f(g(x))·g'(x) dx = ∫ from u = g(a) to u = g(b) f(u) du

When you change the limits, do not substitute back to x. If you keep the original limits, remember they are x-values, so you must rewrite the antiderivative in terms of x before evaluating.

改变上下限后，就不需要再代回 x。若保留原上下限，请记住它们是 x 值，因此必须将原函数用 x 表示后再代入计算。

4. Integration by Parts | 分部积分法

Integration by parts is derived from the product rule for differentiation. It is used when the integrand is a product of two functions, one of which becomes simpler when differentiated and the other is easy to integrate.

分部积分法由微分的乘法法则推导而来。当被积函数是两个函数的乘积，其中一个求导后变简单，另一个容易积分时使用。

∫ u (dv/dx) dx = u v – ∫ v (du/dx) dx

For example, to evaluate ∫ x·eˣ dx, let u = x and dv/dx = eˣ. Then du/dx = 1 and v = eˣ. The formula gives ∫ x eˣ dx = x eˣ – ∫ eˣ dx = x eˣ – eˣ + C = eˣ(x – 1) + C.

例如，计算 ∫ x·eˣ dx 时，令 u = x，dv/dx = eˣ。则 du/dx = 1，v = eˣ。公式给出 ∫ x eˣ dx = x eˣ – ∫ eˣ dx = x eˣ – eˣ + C = eˣ(x – 1) + C。

In the Edexcel formula booklet, integration by parts is not always given, so you must learn it. The key is to choose u and dv carefully before applying the formula.

在 Edexcel 公式手册中，分部积分公式不一定提供，因此必须牢记。关键在于应用公式前仔细选择 u 和 dv。

5. Choosing u and dv: The LIATE Rule | 选择 u 和 dv：LIATE 法则

The order of preference for u is often: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. This helps you choose u so that du is simpler than u and dv can be integrated easily.

选择 u 的优先顺序通常为：对数函数、反三角函数、代数函数、三角函数、指数函数。这有助于选择 u 使其导函数更简单，并确保 dv 容易积分。

Logarithmic functions such as ln x are almost always chosen as u because their derivatives are simpler.
Inverse trig functions such as arctan x are also good choices for u.
Algebraic functions like x or x² are chosen as u when multiplied by exponential or trig functions.

对数函数如 ln x 几乎总是选作 u，因为其导数更简单。
反三角函数如 arctan x 也适合选作 u。
代数函数如 x 或 x² 在与指数函数或三角函数相乘时常选作 u。

For example, in ∫ x·ln x dx, choose u = ln x and dv/dx = x. Then du/dx = 1/x and v = x²/2. The integral becomes (x²/2)ln x – ∫ (x²/2)·(1/x) dx = (x²/2)ln x – x²/4 + C.

例如，在 ∫ x·ln x dx 中，选择 u = ln x，dv/dx = x。则 du/dx = 1/x，v = x²/2。积分变为 (x²/2)ln x – ∫ (x²/2)·(1/x) dx = (x²/2)ln x – x²/4 + C。

6. Repeated Integration by Parts | 重复分部积分法

Sometimes you need to apply integration by parts more than once. This happens with integrals like ∫ x² sin x dx or ∫ eˣ cos x dx, where the original integral reappears after two steps.

有时需要多次使用分部积分法。例如 ∫ x² sin x dx 或 ∫ eˣ cos x dx，经过两步后原积分会重新出现，此时可移项求解。

Let I = ∫ eˣ cos x dx. After two integrations by parts, I = eˣ sin x + eˣ cos x – I ⇒ 2I = eˣ(sin x + cos x) ⇒ I = ½ eˣ(sin x + cos x) + C

When the original integral reappears, add it to both sides instead of cancelling. Many students forget this and lose the factor of 1/2 or a similar constant.

当原积分重新出现时，应将其加到等式两边，而不是抵消。许多学生忘记这一点，导致丢失 1/2 或类似的系数。

For ∫ x² sin x dx, you would apply parts twice, reducing x² to 2x and then to 2. Each step must be written clearly to earn method marks.

对于 ∫ x² sin x dx，需要分部积分两次，将 x² 降为 2x，再降为 2。每一步都要书写清晰才能获得方法分。

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

When the integrand is a rational function where the denominator can be factorised, express it as partial fractions. Each fraction then integrates to a natural logarithm or a standard inverse trig function.

当被积函数是有理函数且分母可以因式分解时，将其表示为部分分式。每个分式可积分为自然对数或标准反三角函数。

∫ 1/(x – a) dx = ln|x – a| + C

For example, to integrate (3x + 5)/[(x – 1)(x + 2)], write it as A/(x – 1) + B/(x + 2). Multiplying through gives 3x + 5 = A(x + 2) + B(x – 1). Setting x = 1 gives A = 8/3, and setting x = -2 gives B = 1/3.

例如，对 (3x + 5)/[(x – 1)(x + 2)] 进行积分，先将其写成 A/(x – 1) + B/(x + 2)。通分后得到 3x + 5 = A(x + 2) + B(x – 1)。令 x = 1 得 A = 8/3，令 x = -2 得 B = 1/3。

The integral is therefore (8/3)ln|x – 1| + (1/3)ln|x + 2| + C. In Edexcel exams, the denominator is often already factorised, but you may need to factorise a quadratic first.

因此积分为 (8/3)ln|x – 1| + (1/3)ln|x + 2| + C。在 Edexcel 考试中，分母通常已分解，但有时需要先对二次式进行因式分解。

8. Selecting the Right Method | 选择正确的方法

In the exam, you must quickly decide which technique to use. Look for composite functions (substitution), products (parts), or rational functions with factorisable denominators (partial fractions).

在考试中，必须快速决定使用哪种技巧。看到复合函数考虑换元法，看到乘积考虑分部积分法，看到分母可分解的有理函数考虑部分分式法。

Integral form	Suggested method
∫ f(g(x))·g'(x) dx	Substitution
∫ xⁿ eᵃˣ dx, ∫ xⁿ sin ax dx	Integration by parts
∫ rational function / quadratic	Partial fractions

A single integral may require two techniques, such as a substitution followed by partial fractions, but Edexcel usually guides you with part (a) and part (b).

一个积分可能需要两种技巧，例如先换元再使用部分分式，但 Edexcel 通常会用第 (a) 问和第 (b) 问来引导你。

9. Common Exam Mistakes | 常见考试错误

Many students forget to change limits in definite substitution, or they mix up signs when integrating by parts. Always write down u, du, dv and v clearly, and check the final answer by differentiation.

许多学生在定积分换元时忘记改变上下限，或在分部积分时符号出错。务必清晰写出 u、du、dv 和 v，并通过求导检验最终答案。

Another common error is losing the constant of integration after finding an indefinite integral. In differential equation contexts, this can prevent you from applying boundary conditions correctly.

另一个常见错误是在求出不定积分后丢失积分常数。在微分方程情境中，这可能导致无法正确应用边界条件。

When using partial fractions, do not forget the absolute value signs in ln|x – a|. The logarithm of a negative number is undefined in real numbers, so the modulus is essential.

使用部分分式时，不要忘记 ln|x – a| 中的绝对值符号。负数的对数在实数范围内无定义，因此取模是必要的。

10. Worked Example: Mixed Practice | 例题：混合练习

Worked example combining substitution and partial fractions: evaluate ∫ (2x + 1)/(x² + x – 2) dx. First factorise the denominator as (x + 2)(x – 1). Then write the integrand as A/(x + 2) + B/(x – 1).

结合换元法和部分分式的例题：计算 ∫ (2x + 1)/(x² + x – 2) dx。首先将分母因式分解为 (x + 2)(x – 1)。然后将被积函数写成 A/(x + 2) + B/(x – 1)。

2x + 1 = A(x – 1) + B(x + 2)

Setting x = 1 gives 3 = 3B, so B = 1. Setting x = -2 gives -3 = -3A, so A = 1. The integral becomes ∫ 1/(x + 2) dx + ∫ 1/(x – 1) dx = ln|x + 2| + ln|x – 1| + C.

令 x = 1 得 3 = 3B，因此 B = 1。令 x = -2 得 -3 = -3A，因此 A = 1。积分变为 ∫ 1/(x + 2) dx + ∫ 1/(x – 1) dx = ln|x + 2| + ln|x – 1| + C。

This can be simplified to ln|(x + 2)(x – 1)| + C, which is equivalent to ln|x² + x – 2| + C. Always check by differentiating to see if you recover the original integrand.

该结果可简化为 ln|(x + 2)(x – 1)| + C，即 ln|x² + x – 2| + C。务必通过求导检验是否能还原为原被积函数。

11. Summary | 内容总结

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Integration Techniques for Edexcel A-Level Pure 4 | Edexcel A-Level Pure 4 积分技巧