B8i Part 4: Advanced Integration and Differential Equations | B8i 第4部分：高级积分与微分方程

📚 B8i Part 4: Advanced Integration and Differential Equations | B8i 第4部分：高级积分与微分方程

In this part of the B8i revision series, we focus on advanced integration techniques and first-order and second-order differential equations that frequently appear in A-level Mathematics papers.

在 B8i 复习系列的这一部分中，我们重点讲解高级积分技巧以及一阶和二阶微分方程，这些都是 A-level 数学考试中的高频考点。

1. Overview of B8i Part 4 | B8i 第4部分概览

This unit connects integration techniques with differential equations. You should be able to identify which method is suitable for a given integral and translate a physical statement into a differential equation.

本单元将积分技巧与微分方程联系起来。你应该能够识别对于给定积分适用哪种方法，并将物理陈述转化为微分方程。

The core skills are substitution, integration by parts, partial fractions, separation of variables, and the integrating factor method.

核心技能包括换元法、分部积分法、部分分式法、变量分离法和积分因子法。

dy/dx = f(x)g(y) ⇒ ∫ 1/g(y) dy = ∫ f(x) dx

2. Integration by Substitution | 换元积分法

Use substitution when an integrand contains a function and its derivative, or when a simple change of variable simplifies a composite expression.

当被积函数包含一个函数及其导数，或者简单的变量替换能够简化复合表达式时，使用换元法。

Choose u = g(x), find du/dx = g'(x), rewrite the integral in terms of u, integrate, and substitute back.

选择 u = g(x)，求出 du/dx = g'(x)，用 u 重写积分，积分后再代回原变量。

∫ f(g(x)) g'(x) dx = ∫ f(u) du

Example: ∫ 2x e^(x²) dx. Let u = x², then du/dx = 2x, so du = 2x dx. The integral becomes ∫ eᵘ du = eᵘ + C = e^(x²) + C.

示例：∫ 2x e^(x²) dx。令 u = x²，则 du/dx = 2x，所以 du = 2x dx。积分变为 ∫ eᵘ du = eᵘ + C = e^(x²) + C。

3. Integration by Parts | 分部积分法

Integration by parts is used for products of functions, such as polynomial times exponential, logarithmic, or trigonometric functions.

分部积分法适用于函数的乘积，例如多项式乘以指数函数、对数函数或三角函数。

Use the formula ∫ u dv = uv − ∫ v du. Choose u according to the LIATE order: logarithmic, inverse trig, algebraic, trigonometric, exponential.

使用公式 ∫ u dv = uv − ∫ v du。根据 LIATE 优先顺序选择 u：对数函数、反三角函数、代数函数、三角函数、指数函数。

∫ u dv = uv − ∫ v du

Example: ∫ x ln x dx. Let u = ln x, dv = x dx, then du = (1/x) dx, v = x²/2. The integral becomes (x²/2) ln x − ∫ (x²/2)(1/x) dx = (x²/2) ln x − x²/4 + C.

示例：∫ x ln x dx。令 u = ln x，dv = x dx，则 du =

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