B9i Part 4: Advanced Integration Techniques and Applications | B9i 第4部分：进阶积分技巧与应用

📚 B9i Part 4: Advanced Integration Techniques and Applications | B9i 第4部分：进阶积分技巧与应用

Integration is one of the most powerful tools in A-Level Mathematics. In Part 4, we focus on advanced techniques such as integration by parts, substitution, partial fractions, parametric integration, volumes of revolution, and first-order differential equations. These methods frequently appear in pure mathematics papers and provide the foundation for university-level calculus.

积分是 A-Level 数学中最强大的工具之一。第4部分重点讲解分部积分、换元积分、部分分式积分、参数方程积分、旋转体积以及一阶微分方程等进阶方法。这些方法在纯数学试卷中经常出现，也是大学微积分的基础。

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1. Core Standard Integrals Review | 核心标准积分回顾

Before applying any advanced method, you must be able to recognise the standard integrals instantly. These results are assumed in most A-Level pure mathematics markschemes.

在应用任何进阶方法之前，你必须能够立即识别标准积分。大多数 A-Level 纯数学评分方案都默认考生掌握这些结果。

The power rule is the first result to memorise:

幂函数积分法则是第一个要熟记的结果：

∫ xⁿ dx = xⁿ⁺¹/(n+1) + C for n ≠ −1

nent. The special case n = −1 gives a logarithm because the denominator would become zero.

当 n ≠ −1 时，幂函数积分法则将指数增加 1，再除以新的指数。特殊情形 n = −1 会得到对数，因为分母会变为零。

∫ x⁻¹ dx = ln|x| + C

You should also recall the exponential and trigonometric standard integrals. These appear repeatedly inside substitution and integration by parts.

你还应该熟记指数函数和三角函数的标准积分。它们在换元积分和分部积分中反复出现。

Function	Integral
eˣ	eˣ + C
sin x	−cos x + C
cos x	sin x + C
sec² x	tan x + C

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2. Integration by Parts | 分部积分法

Integration by parts reverses the product rule for differentiation. The formula is:

分部积分法逆用了微分的乘法法则。其公式为：

∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx

This method is especially useful when the integrand is a product of two functions from different families, such as a polynomial multiplied by an exponential, a logarithm, or a trigonometric function.

当被积函数是两类不同函数族的乘积时，该方法特别有用，例如多项式乘以指数函数、对数函数或三角函数。

A common strategy is to choose u so that du/dx is simpler than u, and choose dv/dx so that v can be found easily. The acronym LIATE helps: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Choose u according to this order.

常用策略是选取 u 使得 du/dx 比 u 更简单，同时选取 dv/dx 使 v 容易求得。LIATE 助记法会很有帮助：对数函数、反三角函数、代数函数、三角函数、指数函数。按照这个顺序选择 u。

Worked example: find ∫ x eˣ dx. Let u = x and dv/dx = eˣ. Then du/dx = 1 and v = eˣ.

例题：求 ∫ x eˣ dx。令 u = x，dv/dx = eˣ，则 du/dx = 1，v = eˣ。

∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C

es ln x, let u = ln x because its derivative 1/x simplifies the calculation significantly.

如果乘积中包含 ln x，通常令 u = ln x，因为其导数 1/x 会显著简化计算。

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3. Integration by Substitution | 换元积分法

Substitution reverses the chain rule. The general principle is:

换元积分法逆用了链式法则。一般原理是：

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

For definite integrals, you must change the limits when changing the variable. If x goes from a to b, then u goes from g(a) to g(b). This avoids the need to substitute back at the end.

对于定积分，换元后必须同步替换积分限。如果 x 从 a 变到 b，那么 u 从 g(a) 变到 g(b)。这样就不需要在最后再把变量换回来。

Worked example: find ∫ 2x e^x² dx. Let u = x², so du/dx = 2x and du = 2x dx. The integral becomes ∫ eᵘ du = eᵘ + C.

例题：求 ∫ 2x e^x² dx。令 u = x²，所以 du/dx = 2x，du = 2x dx。原积分变为 ∫ eᵘ du = eᵘ + C。

∫ 2x e^x² dx = e^x² + C

When choosing u, look for an inner function whose derivative appears as a factor in the integrand. Common choices include u inside a bracket, under a root, in a denominator, or as an exponent.

选择 u 时，寻找一个内层函数，其导数作为因子出现在被积函数中。常见的选择包括括号内、根号下、分母中或指数位置上的函数。

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4. Partial Fractions | 部分分式积分

When integrating rational functions, first check whether the denominator factorises. If the degree of the numerator is less than the degree of the denominator, split into partial fractions.

在积分有理函数时，首先检查分母是否可以因式分解。如果分子的次数低于分母的次数，就拆分成部分分式。

Worked example: find ∫ (2x+1)/((x−1)(x+2)) dx. Write the fraction as A/(x−1) + B/(x+2).

例题：求 ∫ (2x+1)/((x−1)(x+2)) dx。将该分式写成 A/(x−1) + B/(x+2)。

2x+1 = A(x+2) + B(x−1)

Set x = 1 to find A: 3 = 3A, so A = 1. Set x = −2 to find B: −3 = −3B, so B = 1.

令 x = 1 求 A：3 = 3A，得到 A = 1。令 x = −2 求 B：−3 = −3B，得到 B = 1。

∫ 1/(x−1) dx + ∫ 1/(x+2) dx = ln|x−1| + ln|x+2| + C

If the denominator has a repeated linear factor such as (x−1)², include both A/(x−1) and B/(x−1)² in the decomposition.

如果分母含有重复线性因子，例如 (x−1)²，分解时必须同时包含 A/(x−1) 和 B/(x−1)²。

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5. Parametric Integration | 参数方程积分

ve is given by x = x(t) and y = y(t), the area under the curve between t = a and t = b is:

如果曲线由 x = x(t) 和 y = y(t) 给出，则曲线在 t = a 与 t = b 之间下方的面积为：

A = ∫ y dx = ∫ y(t) x′(t) dt

The key idea is to replace dx with x′(t) dt and then change the limits from x-values to t-values. This makes the integral purely in terms of the parameter t.

核心思想是用 x′(t) dt 替换 dx，并将积分限从 x 值转换为 t 值。这样积分就完全用参数 t 表示。

or example, if x = t² and y = t³, then dx/dt = 2t, so the area is ∫ t³ × 2t dt = ∫ 2t⁴ dt.

例如，如果 x = t² 且 y = t³，则 dx/dt = 2t，因此面积为 ∫ t³ × 2t dt = ∫ 2t⁴ dt。

This technique is also used when finding arc lengths or surface areas in further mathematics. At A-Level, the main application is area under a parametric curve.

该技巧在进阶数学中也用于求弧长或曲面面积。在 A-Level 阶段，主要应用是求参数曲线下方的面积。

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6. Volumes of Revolution | 旋转体积

When a region is rotated about the x-axis, the volume is given by:

当一个区域绕 x 轴旋转时，其体积为：

V = π ∫ y² dx

For rotation about the y-axis, the volume is:

绕 y 轴旋转时，体积为：

V = π ∫ x² dy

It is essential to square the correct variable before integrating. Many exam errors happen because students forget the square or use the wrong axis.

在积分之前必须对正确的变量进行平方。许多考试错误是因为学生忘记平方，或者使用了错误的轴。

Worked example: rotate y = √x from x = 0 to x = 4 about the x-axis. Then y² = x, so V = π ∫ x dx from 0 to 4.

例题：将 y = √x 从 x = 0 到 x = 4 绕 x 轴旋转。则 y² = x，因此 V = π ∫ x dx，积分限为 0 到 4。

V = π [x²/2] from 0 to 4 = π(16/2 − 0) = 8π

For composite volumes, subtract the inner hollow volume from the outer solid volume. This is common when a region between two curves is rotated.

对于复合体积，

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