Integration by Parts | 分部积分法

📚 Integration by Parts | 分部积分法

In A-Level Edexcel Pure Mathematics, integration by parts is a key technique for integrating products of two functions. It often appears in Paper 2 and Paper 3 questions involving exponentials, logarithms, trigonometric functions, and sometimes repeated applications.

在 A-Level Edexcel 纯数学中，分部积分法是积分两个函数乘积的关键技巧。它常出现在 Paper 2 和 Paper 3 的题目中，涉及指数函数、对数函数、三角函数，有时还需要重复使用。

This method is fundamentally the reverse of the product rule for differentiation, and it is especially useful when one factor becomes simpler after differentiation while the other is straightforward to integrate.

该方法本质上是微分乘积法则的逆运算，当一个因子在微分后变得更简单，而另一个因子容易积分时，它尤其有用。

1. The Core Formula | 核心公式

For two differentiable functions u = u(x) and v = v(x), the integration by parts formula states that:

对于两个可导函数 u = u(x) 和 v = v(x)，分部积分公式为：

∫ u dv/dx dx = uv − ∫ v du/dx dx

This is often remembered as ‘uv minus the integral of v du’. The formula works by transferring the derivative from one function to the other, which can turn a difficult integral into an easier one.

这通常记作 “uv 减去 v du 的积分”。该公式通过将导数从一个函数转移到另一个函数，可以把困难的积分转化为更容易的积分。

The symbol u and v are not fixed; you choose them from the two factors in the original product. This choice is the most important strategic decision in using the technique correctly.

符号 u 和 v 并不是固定不变的；你从原乘积的两个因子中选择它们。这一选择是正确使用该技巧的最重要策略。

2. Derivation from the Product Rule | 由乘积法则推导

The product rule for differentiation states that:

微分的乘积法则为：

d/dx (uv) = u dv/dx + v du/dx

Rearranging this equation gives u dv/dx = d/dx (uv) − v du/dx. This is exactly the form needed before integrating both sides with respect to x.

移项得 u dv/dx = d/dx (uv) − v du/dx。这正是两边对 x 积分前所需的形式。

Integrating both sides produces the standard formula, because the antiderivative of the derivative of uv is simply uv. This explains why the first term on the right has no integral sign.

对两边积分即可得到标准公式，因为 uv 导数的原函数就是 uv。这解释了为什么右边的第一项没有积分号。

This derivation also shows why the constant of integration is usually added at the end rather than inside intermediate steps. Adding it at the end avoids unnecessary algebraic complications.

这一推导也说明了为什么积分常数通常在最后添加而不是在中间步骤中添加。最后加上常数可以避免不必要的代数复杂化。

3. How to Choose u and dv/dx: The LIATE Rule | 如何选择 u 与 dv/dx：LIATE 法则

A key skill is selecting which factor to call u and which factor to call dv/dx. A useful mnemonic is LIATE: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Choose u as the first type that appears in this list.

关键技巧是选择哪个因子设为 u，哪个设为 dv/dx。一个实用助记符是 LIATE：对数函数、反三角函数、代数函数、三角函数、指数函数。选择列表中先出现的类型为 u。

For example, in ∫ x eˣ dx, x is algebraic and eˣ is exponential, so choose u = x and dv/dx = eˣ. In ∫ ln x dx, choose u = ln x and dv/dx = 1 because logarithmic functions come first in the list.

例如，在 ∫ x eˣ dx 中，x 是代数函数，eˣ 是指数函数，因此选择 u = x，dv/dx = eˣ。在 ∫ ln x dx 中，选择 u = ln x，dv/dx = 1，因为对数函数在列表中排最前。

Order	Function type	Examples
1	Logarithmic \| 对数函数	ln x, (ln x)²
2	Inverse trigonometric \| 反三角函数	arcsin x, arctan x
3	Algebraic \| 代数函数	x, x², 3x + 2
4	Trigonometric \| 三角函数	sin x, cos x, tan x
5	Exponential \| 指数函数	eˣ, e²ˣ, 2ˣ

Choosing the wrong u may still produce a valid equation, but it often makes the new integral more complicated rather than simpler. In such cases, swap the roles and try again.

选择错误的 u 可能仍然得到有效等式，但通常会使新积分变得更复杂而不是更简单。在这种情况下，可以交换角色再试一次。

4. Standard Example: x eˣ | 标准例题：x eˣ

Evaluate ∫ x eˣ dx. Let u = x and dv/dx = eˣ. Then du/dx = 1 and v = eˣ.

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

∫ x eˣ dx = x eˣ − ∫ eˣ × 1 dx = x eˣ − eˣ + C

The result can be written as eˣ(x − 1) + C. This single application is enough because the remaining integral ∫ eˣ dx is a basic exponential integral.

结果可写作 eˣ(x − 1) + C。一次分部积分就足够了，因为剩余的积分 ∫ eˣ dx 是基本指数积分。

This example illustrates the core idea: differentiating x gives a constant, while integrating eˣ leaves it unchanged. Therefore the product becomes simpler after one step.

这个例子说明了核心思想：对 x 微分得到常数，而对 eˣ 积分保持不变。因此乘积经过一步后就变得更简单。

5. Integrating the Natural Logarithm | 积分自然对数

For integrals involving the natural logarithm, set u = ln x and dv/dx = 1. This gives du/dx = 1/x and v = x.

对于涉及自然对数的积分，令 u = ln x，dv/dx = 1。由此 du/dx = 1/x，v = x。

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

For powers of ln x, such as ∫ (ln x)² dx, apply the same idea with u = (ln x)². After one step the remaining integral contains ln x, so repeat the technique.

对于 ln x 的幂，例如 ∫ (ln x)² dx，使用相同的思路令 u = (ln x)²。一步后剩余积分含有 ln x，因此重复该技巧。

∫ (ln x)² dx = x (ln x)² − 2x ln x + 2x + C

This pattern produces a reduction-type effect where the power of the logarithm decreases each time. It is a common exam task because it tests both selection strategy and accuracy.

这种模式产生类似递推的效果，对数的幂每次减少。这是一个常见的考试任务，因为它既考察选择策略也考察准确性。

6. Repeated Integration by Parts | 多次分部积分

When an integral contains x² eˣ or x² sin x, one application is not enough. Apply integration by parts repeatedly, reducing the power of x each time until the remaining integral is basic.

当积分含有 x² eˣ 或 x² sin x 时，一次分部积分不够。需要重复使用分部积分，每次降低 x 的幂，直到剩余积分为基本积分。

For ∫ x² eˣ dx, first let u = x², dv/dx = eˣ, giving v = eˣ and du/dx = 2x. This produces a new integral ∫ 2x eˣ dx, which still requires the same technique.

对于 ∫ x² eˣ dx，首先令 u = x²，dv/dx = eˣ，得到 v = eˣ，du/dx = 2x。这会产生一个新的积分 ∫ 2x eˣ dx，仍然需要相同的技巧。

∫ x² eˣ dx = x² eˣ − ∫ 2x eˣ dx = x² eˣ − 2(x eˣ − eˣ) + C = eˣ(x² − 2x + 2) + C

Careful bookkeeping of signs and coefficients is essential in repeated applications. A single lost negative sign can change the entire final answer.

在多次应用时，必须仔细处理符号和系数。漏掉一个负号就可能导致最终答案完全不同。

7. Cyclic Integration: eˣ cos x | 循环积分：eˣ cos x

For integrals such as ∫ eˣ cos x dx or ∫ eˣ sin x dx, neither repeated differentiation of eˣ nor of the trigonometric function leads to a simpler product. Instead, apply integration by parts twice and solve for the original integral.

对于 ∫ eˣ cos x dx 或 ∫ eˣ sin x dx

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Integration by Parts | 分部积分法