📚 B8i Part 7: Integration by Parts | 分部积分法
Integration by parts is one of the main techniques in A-Level Mathematics for integrating products of functions. It is derived from the product rule for differentiation and is especially useful when the integrand is a product of two different types of functions, such as a polynomial multiplied by an exponential, a logarithm multiplied by a polynomial, or a trigonometric function multiplied by an exponential.
分部积分法是 A-Level 数学中处理函数乘积积分的主要方法之一。它来源于微分的乘积法则，尤其适用于被积函数由两种不同类型的函数相乘组成的情况，例如多项式乘指数函数、对数函数乘多项式、或三角函数乘指数函数等。
1. The Idea Behind Integration by Parts | 分部积分的基本思想
e rearrange this formula and integrate both sides with respect to x, we obtain a rule that allows us to transfer differentiation from one factor to another inside an integral.
微分的乘积法则指出，如果 u 和 v 都是 x 的函数，那么它们乘积的导数为 d/dx (uv) = u dv/dx + v du/dx。如果我们重新整理这个公式，并对两边关于 x 积分，就能得到一条规则，使我们可以把积分中对一个因子的微分作用转移到另一个因子上。
∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx
This formula is the foundation of integration by parts. The key idea is to choose u and dv carefully so that the new integral on the right-hand side is easier to evaluate than the original integral on the left-hand side.
这个公式是分部积分法的基础。核心思想是谨慎地选择 u 和 dv，使得右边的新积分比左边的原积分更容易计算。
2. The Formula and Notation | 公式与记号
is the part we choose to differentiate, dv is the part we choose to integrate, du is the derivative of u, and v is the integral of dv.
在更紧凑的记号中，我们通常把该规则写成 ∫ u dv = uv − ∫ v du。这里 u 是我们选择进行微分的部分，dv 是我们选择进行积分的部分，du 是 u 的导数，v 是 dv 的积分。
du/dx) dx. This version is helpful because it shows explicitly that both u and v are functions of x, and that the integration is with respect to x.
对于 x 的函数，该公式通常写成 ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx。这种形式很有帮助，因为它明确表示 u 和 v 都是 x 的函数，并且积分是对 x 进行的。
∫ u dv = uv − ∫ v du
When using this notation, always remember to include the constant of integration at the end of an indefinite integral. It is a common mistake to omit + C after applying the formula.
使用这种记号时，务必记住在不定期积分的最后加上积分常数。应用公式后漏写 + C 是一个常见错误。
3. Choosing u and dv: The LIATE Rule | 选择 u 和 dv：LIATE 法则
The success of integration by parts depends heavily on choosing u and dv wisely. A useful order of priority for selecting u is known as the LIATE rule, where u is chosen from the category that appears first in the list.
分部积分能否成功，很大程度上取决于是否明智地选择 u 和 dv。一个常用的优先顺序称为 LIATE 法则，即从列表中最先出现的类别中选择 u。
| Priority | Function Type | Example |
|---|---|---|
| L | Logarithmic functions | ln x, ln(2x) |
| I | Inverse trigonometric functions | arctan x, arcsin x |
| A | Algebraic functions | x, x², x³ |
| T | Trigonometric functions | sin x, cos x |
| E | Exponential functions | eˣ, e²ˣ |
For example, in ∫ x eˣ ∫ x ln x dx, the logarithmic factor ln x should be chosen as u because L appears before A.
例如，在 ∫ x eˣ dx 中，代数因子 x 应选为 u，因为在 LIATE 中 A 出现在 E 之前。在 ∫ x ln x dx 中，对数因子 ln x 应选为 u，因为 L 出现在 A 之前。
4. First Worked Example: ∫ x eˣ dx | 第一个例题：∫ x eˣ dx
Evaluate ∫ x eˣ dx. According to the LIATE rule, choose u = x and dv = eˣ dx. Then du = dx and v = eˣ. Substituting into the formula gives ∫ x eˣ dx = x eˣ − ∫ eˣ dx.
计算 ∫ x eˣ dx。根据 LIATE 法则，选择 u = x，dv = eˣ dx。那么 du = dx，v = eˣ。代入公式得到 ∫ x eˣ dx = x eˣ − ∫ eˣ dx。
∫ x eˣ dx = x eˣ − eˣ + C
The new integral ∫ eˣ dx is simply eˣ, so the final answer is x eˣ − eˣ e shows how integration by parts replaces a product integral with a much simpler integral.
新的积分 ∫ eˣ dx 就是 eˣ，因此最终答案为 x eˣ − eˣ + C。这个例子说明分部积分如何把一个乘积积分替换成一个简单得多的积分。
5. Integrating Logarithms and Inverse Trigonometric Functions | 对数和反三角函数的积分
, such as ∫ ln x dx. In this case, let u = ln x and dv = dx. Then du = (1/x) dx and v = x.
分部积分法也可以用来积分看似不是乘积的函数，例如 ∫ ln x dx。此时，令 u = ln x，dv = dx。那么 du = (1/x) dx，v = x。
∫ ln x dx = x ln x − ∫ x (1/x) dx = x ln x − x + C
²) dx. The remaining integral can be evaluated using substitution, leading to (1/2)ln(1 + x²).
对于 ∫ arctan x dx，选择 u = arctan x，dv = dx。那么 du = 1/(1 + x²) dx，v = x。公式给出 x arctan x − ∫ x/(1 + x²) dx。剩下的积分可以用代入法求解，结果为 (1/2)ln(1 + x²)。
∫ arctan x dx = x arctan x − (1/2)ln(1 + x²) + C
These standard results are frequently tested in A-Level exams, so it is useful to remember the method rather than just the final answers.
这些标准结果在 A-Level 考试中经常出现，因此记住方法而不仅仅是最终答案会更有帮助。
6. Repeated Integration by Parts | 重复分部积分
metimes a single application of integration by parts is not enough. For example, in ∫ x² eˣ dx, choosing u = x² and dv = eˣ dx gives du = 2x dx and v = eˣ. The first step produces ∫ x² eˣ dx = x² eˣ − 2 ∫ x eˣ dx.
有时只应用一次分部积分是不够的。例如，在 ∫ x² eˣ dx 中，选择 u = x²，dv = eˣ dx，得到 du = 2x dx 和 v = eˣ。第一步得到 ∫ x² eˣ dx = x² eˣ − 2 ∫ x eˣ dx。
The remaining integral ∫ x eˣ dx still requires integration by parts. From the previous example, we know that ∫ x eˣ dx = x eˣ − eˣ. Substituting this result back gives the final answer.
剩下的积分 ∫ x eˣ dx 仍需分部积分。由前面的例子可知 ∫ x eˣ dx = x eˣ − eˣ。把这个结果代回去即可得到最终答案。
∫ x² eˣ dx = x² eˣ − 2x eˣ + 2eˣ + C
The general pattern for ∫ xⁿ eˣ becomes a constant. This is a powerful technique for many polynomial-exponential products.
对于 ∫ xⁿ eˣ dx，可以通过不断对多项式部分求导直到其变为常数，从而逐步建立一般规律。这是处理许多多项式与指数函数乘积问题的有力技巧。
7. Definite Integrals | 定积分
m is evaluated at the upper and lower limits. The formula is ∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du.
对于定积分，分部积分使用相同的公式，但 uv 这一项要代入上下限进行求值。公式为 ∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du。
∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du
Example: Evaluate ∫₀¹ x eˣ dx. Choose u = x and dv = eˣ dx, so du = dx and v = eˣ. Applying the formula gives [x eˣ]₀¹ − ∫₀¹ eˣ dx = (1 × e¹ − 0) − [eˣ]₀¹ = e − (e − 1) = 1.
例题：计算 ∫₀¹ x eˣ dx。选择 u = x，dv = eˣ dx，所以 du = dx，v = eˣ。应用公式得到 [x eˣ]₀¹ − ∫₀¹ eˣ dx = (1 × e¹ − 0) − [eˣ]₀¹ = e − (e − 1) = 1。
When handling definite integrals, it is important to apply the limits carefully to the uv term before subtracting the remaining integral. A common error is to forget the limits on the second integral.
处理定积分时，务必先把上下限代入 uv 项，再减去剩下的积分。一个常见错误是忘记第二个积分也要带上下限。
8. The Boomerang Method | 回旋镖法
the original integral after applying integration by parts twice. A classic example is ∫ eˣ sin x dx. Let I = ∫ eˣ sin x dx. First choose u = sin x and dv = eˣ dx, giving du = cos x dx and v = eˣ. This produces I = eˣ sin x − ∫ eˣ cos x dx.
有些积分在两次分部积分后会回到原来的积分。一个经典例子是 ∫ eˣ sin x dx。设 I = ∫ eˣ sin x dx。首先选择 u = sin x，dv = eˣ dx，得到 du = cos x dx 和 v = eˣ。于是 I = eˣ sin x − ∫ eˣ cos x dx。
Now apply integration by parts to ∫ eˣ cos x dx. Choose u = cos x and dv = eˣ dx, giving du = −sin x dx and v = eˣ. The result is ∫ eˣ cos x dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I.
现在对 ∫ eˣ cos x dx 应用分部积分。选择 u = cos x，dv = eˣ dx，得到 du = −sin x dx 和 v = eˣ。结果是 ∫ eˣ cos x dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I。
Substitute this back into the first expression: I = eˣ sin x − (eˣ cos x + I) = eˣ sin x − eˣ cos x − I. Rearranging gives 2I = eˣ sin x − eˣ cos x, so I = (1/2)eˣ(sin x − cos x) + C.
把这个结果代回第一个表达式：I = eˣ sin x − (eˣ cos x + I) = eˣ sin x − eˣ cos x − I。重新整理得到 2I = eˣ sin x − eˣ cos x，因此 I = (1/2)eˣ(sin x − cos x) + C。
∫ eˣ sin x dx = (1/2)eˣ(sin x − cos x) + C
This method is called the boomerang method because the original integral returns after two steps, allowing it to be solved algebraically rather than by repeated integration.
这种方法称为回旋镖法，因为原积分在两步之后回到等式右边，从而可以通过代数方法求解，
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