A-Level Maths B8i Part 5: Advanced Integration Techniques | A-Level数学B8i第5部分：高级积分技巧

📚 A-Level Maths B8i Part 5: Advanced Integration Techniques | A-Level数学B8i第5部分：高级积分技巧

Advanced integration sits at the heart of A-level pure mathematics. In this part, we build on basic antiderivatives and develop the main techniques you need for exam success: substitution, integration by parts, partial fractions, trigonometric integrals, and applications such as areas and volumes of revolution.

高级积分是A-level纯数学的核心内容。在本部分中，我们在基本不定积分的基础上，逐步掌握考试成功所需的主要技巧：换元积分、分部积分、部分分式、三角积分，以及面积与旋转体体积等应用。

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1. Core Integration Rules Review | 核心积分法则回顾

Before moving to advanced methods, you must be able to recall the standard results instantly. The fundamental antiderivatives include the power rule, the exponential rule, the natural logarithm rule, and the two basic trigonometric rules.

在学习高级技巧之前，你必须能够快速回忆起标准积分结果。基本的不定积分包括幂函数、指数函数、自然对数以及两个基本三角函数的积分法则。

• ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, for n ≠ −1
• ∫ eˣ dx = eˣ + C
• ∫ 1/x dx = ln |x| + C
• ∫ cos x dx = sin x + C
• ∫ sin x dx = −cos x + C

These are the building blocks. If any of them feels slow, pause and practise until it becomes automatic, because advanced questions always depend on them.

这些是最基础的积木。如果其中任何一个让你觉得不够熟练，请先停下来练习直到能够自动反应，因为高级题目始终依赖于这些基本结果。

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2. Integration by Substitution | 换元积分法

The chain rule for differentiation reverses into the method of substitution. If an integrand contains a composite function f(g(x)) together with g'(x), set u = g(x), replace g'(x)dx by du, and integrate with respect to u. The final answer must be returned to the original variable x.

微分的链式法则反过来就是换元积分法。如果被积函数中包含复合函数 f(g(x)) 以及 g'(x)，就设 u = g(x)，把 g'(x)dx 替换为 du，然后对 u 积分。最后必须把结果换回原来的变量 x。

Example: Evaluate ∫ 2x √(x² + 1) dx.

例题：计算 ∫ 2x √(x² + 1) dx。

Let u = x² + 1, so du/dx = 2x, hence du = 2x dx. The integral becomes ∫ √u du = (2/3)u^(3/2) + C = (2/3)(x² + 1)^(3/2) + C.

设 u = x² + 1，则 du/dx = 2x，因此 du = 2x dx。积分变为 ∫ √u du = (2/3)u^(3/2) + C = (2/3)(x² + 1)^(3/2) + C。

∫ 2x √(x²+1) dx = (2/3)(x²+1)^(3/2) + C

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3. Substitution with Definite Integrals | 换元积分法在定积分中的应用

When a substitution changes the variable in a definite integral, you have two options: rewrite the antiderivative back in x and keep the original limits, or change the limits to the new variable and never go back to x. The second method is usually faster and cleaner.

当换元法改变定积分中的变量时，你有两种选择：把原函数换回 x 并保持原来的上下限，或者把上下限一起换到新变量，不再回到 x。第二种方法通常更快、更简洁。

Example: Evaluate ∫ from 0 to 1 of 2x e^(x²) dx.

例题：计算 ∫ 从 0 到 1 的 2x e^(x²) dx。

Let u = x². Then du = 2x dx. When x = 0, u = 0; when x = 1, u = 1. The integral becomes ∫ from 0 to 1 of e^u du = [e^u] from 0 to 1 = e − 1.

设 u = x²。则 du = 2x dx。当 x = 0 时，u = 0；当 x = 1 时，u = 1。积分变为 ∫ 从 0 到 1 的 e^u du = [e^u] 从 0 到 1 = e − 1。

∫₀¹ 2x e^(x²) dx = e − 1

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4. Integration by Parts | 分部积分法

Integration by parts is the reverse of the product rule. The formula is ∫ u dv = uv − ∫ v du. Choose u according to the LIATE rule: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. This usually selects the right function to differentiate.

分部积分法是乘积法则的逆运算。公式为 ∫ u dv = uv − ∫ v du。按照 LIATE 规则选择 u：对数函数、反三角函数、代数函数、三角函数、指数函数。这个顺序通常能帮助你选对需要求导的函数。

Example: Evaluate ∫ x eˣ dx.

例题：计算 ∫ x eˣ dx。

Let u = x, dv = eˣ dx. Then du = dx, v = eˣ. Hence ∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C.

设 u = x，dv = eˣ dx。则 du = dx，v = eˣ。因此 ∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C。

∫ x eˣ dx = eˣ(x − 1) + C

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5. Repeated Integration by Parts | 分部积分法的多次应用

Some integrals require applying integration by parts two or more times. This often happens with powers of x multiplied by sin x, cos x, or eˣ. After two applications, the original integral may reappear, so you can solve for it algebraically.

有些积分需要连续使用两次或更多次分部积分法。这种情况通常出现在 x 的幂乘以 sin x、cos x 或 eˣ 时。使用两次后，原积分可能会重新出现，这时可以通过代数方法解出原积分。

Example: Evaluate ∫ x² sin x dx.

例题：计算 ∫ x² sin x dx。

First let u = x², dv = sin x dx, so v = −cos x. This gives −x² cos x + ∫ 2x cos x dx. Apply parts again with u = 2x, dv = cos x dx, giving 2x sin x + 2 cos x. Therefore the final result is −x² cos x + 2x sin x + 2 cos x + C.

首先设 u = x²，dv = sin x dx，则 v = −cos x。得到 −x² cos x + ∫ 2x cos x dx。再次使用分部积分，设 u = 2x，dv = cos x dx，得到 2x sin x + 2 cos x。因此最终结果为 −x² cos x + 2x sin x + 2 cos x + C。

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