📚 Integration by Parts | 分部积分法
Integration by parts is a fundamental technique in calculus, especially essential for the IB Mathematics: Analysis and Approaches HL syllabus. It allows us to integrate products of functions by reversing the product rule for differentiation. Mastering this method opens the door to a wide variety of integrals that cannot be tackled by simple substitution alone, including those involving polynomials multiplied by exponentials, trigonometric functions, or logarithms.
分部积分法是微积分中的一项基本技巧，对于 IB 数学：分析与方法 HL 课程尤为重要。它通过逆用微分的乘积法则，让我们能够对函数的乘积进行积分。掌握这一方法，就可以处理许多单纯用换元法无法解决的积分，包括多项式乘以指数函数、三角函数或对数函数等类型。
1. Introduction to Integration by Parts | 分部积分法简介
(x)g(x) dx, and no obvious substitution simplifies it, integration by parts becomes the tool of choice. It transforms the original integral into a (hopefully) simpler one, based on the clever choice of two parts: one to differentiate and the other to integrate. The core idea is to transfer the derivative from one factor to the other, reducing complexity step by step.
当我们面对形如 ∫ f(x)g(x) dx 的积分，并且没有明显的换元可以简化它时，分部积分法就成为了首选工具。它通过巧妙地选择两部分——一部分求导，另一部分积分——把原积分转化为（希望是）更简单的积分。其核心思想是将导数从一个因子转移到另一个因子上，逐步降低复杂度。
2. Derivation from the Product Rule | 从乘积法则推导
and v(x), the derivative of their product is given by (d/dx)(u v) = u (dv/dx) + v (du/dx).
分部积分公式直接由微分的乘积法则推导而来。回想一下，对于两个可微函数 u(x) 和 v(x)，它们乘积的导数为 (d/dx)(u v) = u (dv/dx) + v (du/dx)。
ies to uv = ∫ u dv + ∫ v du. Rearranging gives the classical integration by parts formula:
对两边同时关于 x 积分，我们得到 ∫ d(uv) = ∫ u dv + ∫ v du，化简后为 uv = ∫ u dv + ∫ v du。移项便得到经典的分部积分公式：
∫ u dv = uv − ∫ v du
ppropriately, so that the new integral ∫ v du is easier to evaluate than the original one.
应用这个公式的技巧在于恰当地选择 u 和 dv，使得新的积分 ∫ v du 比原积分更容易计算。
3. The LIATE Rule for Choosing u | 选择 u 的 LIATE 规则
A practical mnemonic to guide the choice of u is the acronym LIATE, which lists function types in order of preference for u: Logarithmic functions, Inverse trigonometric functions, Algebraic (polynomial) functions, Trigonometric functions, and Exponential functions. The rule suggests we should choose u as the function that appears first in this list, because its derivative will typically be simpler, while dv takes the rest.
一个指导选择 u 的实用口诀是缩写 LIATE，它按优先顺序列出了函数类型：L 对数函数、I 反三角函数、A 代数（多项式）函数、T 三角函数、E 指数函数。该规则建议我们选择出现在这个列表中靠前的函数作为 u，因为它的导数通常会变得更简单，而 dv 则包含剩余部分。
For example, in ∫ x eˣ dx, x is algebraic (A) and eˣ is exponential (E); according to LIATE, A comes before E, so we set u = x and dv = eˣ logarithmic (L), so u = ln x, and dv = dx. This systematic approach resolves most choices.
例如，在 ∫ x eˣ dx 中，x 是代数函数 (A)，eˣ 是指数函数 (E)；根据 LIATE，A 排在 E 之前，因此我们令 u = x，dv = eˣ dx。在 ∫ ln x dx 中，可以看作 ∫ (ln x)·1 dx；ln x 是对数函数 (L)，所以令 u = ln x，dv = dx。这种系统的方法能解决大多数的选择。
4. Basic Examples: Polynomial Times Exponential | 基础例题：多项式乘指数函数
Consider the integral ∫ x eˣ dx. Following LIATE, set u = x, so du = dx, and dv = eˣ dx, giving v = eˣ. Substituting into the formula yields:
考虑积分 ∫ x eˣ dx。按照 LIATE 规则，令 u = x，则 du = dx；令 dv = eˣ dx，则 v = eˣ。代入公式得：
∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C
When a higher power of x is present, such as ∫ x² eˣ dx, we need to apply the procedure twice. First set u = x², dv = eˣ dx → du = 2x dx, v = eˣ, giving ∫ x² eˣ dx = x² eˣ − 2∫ x eˣ dx. The remaining integral is the one we just solved, leading to x² eˣ − 2(x eˣ − eˣ) + C = eˣ(x² − 2x + 2) + C. Each application reduces the polynomial degree.
当出现 x 的更高次幂时，比如 ∫ x² eˣ dx，我们需要两次运用该公式。首先令 u = x²，dv = eˣ dx → du = 2x dx，v = eˣ，得到 ∫ x² eˣ dx = x² eˣ − 2∫ x eˣ dx。剩下的积分就是我们刚计算过的，于是得到 x² eˣ − 2(x eˣ − eˣ) + C = eˣ(x² − 2x + 2) + C。每次运用都会降低多项式的次数。
5. Examples with Logarithmic Functions | 对数函数型例题
braic and simpler. For ∫ ln x dx, set u = ln x, dv = dx, so du = (1/x) dx and v = x. Then:
对数函数几乎总是被选作 u，因为它们的导数是代数函数并且更简单。对于 ∫ ln x dx，令 u = ln x，dv = dx，则 du = (1/x) dx，v = x。于是：
∫ ln x dx = x ln x − ∫ x·(1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C
C. The logarithm disappears after differentiation, leaving a pure polynomial to integrate.
类似地，∫ x ln x dx 需要令 u = ln x，dv = x dx → du = (1/x) dx，v = ½ x²。代入公式得到 ½ x² ln x − ∫ ½ x²·(1/x) dx = ½ x² ln x − ½ ∫ x dx = ½ x² ln x − ¼ x² + C。对数函数在求导后消失，留下一个纯多项式来积分。
6. Examples with Inverse Trigonometric Functions | 反三角函数型例题
an x dx. Set u = arctan x, dv = dx → du = 1/(1+x²) dx, v = x. Integration by parts yields:
反三角函数同样优先作为 u。以 ∫ arctan x dx 为例，令 u = arctan x，dv = dx → du = 1/(1+x²) dx，v = x。分部积分得：
∫ arctan x dx = x arctan x − ∫ x/(1+x²) dx
ning integral can be handled with a simple substitution w = 1+x², dw = 2x dx, giving −½ ln|1+x²| pattern shows that integration by parts often reduces the problem to a basic substitution.
剩下的积分可以通过简单的换元 w = 1+x²，dw = 2x dx 来处理，得到 −½ ln|1+x²| + C。因此最终结果为 x arctan x − ½ ln(1+x²) + C。这个模式表明，分部积分法常常将问题化归为一个基本换元。
7. Repeated Integration by Parts | 重复分部积分法
ance, evaluate ∫ x² sin x dx. Choose u = x², dv = sin x dx → du = 2x dx, v = −cos x. Then:
当积分包含多项式乘以三角函数时，我们通常需要多次运用分部积分法，直到多项式次数降为零。例如，计算 ∫ x² sin x dx。选择 u = x²，dv = sin x dx → du = 2x dx，v = −cos x。则：
∫ x² sin x dx = −x² cos x + ∫ 2x cos x dx
x + 2 cos x. Putting it all together: ∫ x² sin x dx = −x² cos x + 2x sin x + 2 cos x + C.
现在对 ∫ 2x cos x dx 再次运用分部积分法，令 u = 2x，dv = cos x dx → du = 2 dx，v = sin x。得到 ∫ 2x cos x dx = 2x sin x − ∫ 2 sin x dx = 2x sin x + 2 cos x。综合起来：∫ x² sin x dx = −x² cos x + 2x sin x + 2 cos x + C。
8. Cyclic Integration (Solving for the Integral) | 循环积分（解方程求积分）
unctions, lead to a cyclic pattern where the original integral reappears. Consider I = ∫ eˣ cos x dx. Let u = cos x, dv = eˣ dx → du = −sin x dx, v = eˣ. Then I = eˣ cos x + ∫ eˣ sin x dx. Now apply integration by parts to the new integral with u = sin x, dv = eˣ dx → du = cos x dx, v = eˣ, giving ∫ eˣ sin x dx = eˣ sin x − ∫ eˣ cos x dx = eˣ sin x − I.
有些积分，尤其是指数函数与三角函数的乘积，会导致循环出现，原积分再次登场。考虑 I = ∫ eˣ cos x dx。令 u = cos x，dv =
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