Integration by Parts | 分部积分法

📚 Integration by Parts | 分部积分法

Integration by parts is a fundamental technique in calculus that allows us to integrate products of functions. Derived from the product rule for differentiation, it transforms a difficult integral into a simpler one by strategically choosing which part to differentiate and which to integrate. In the IB Mathematics: Analysis and Approaches HL course, mastering this method is essential for tackling a wide range of integrals, from polynomials multiplied by exponentials to cyclic trigonometric forms.

分部积分法是微积分中一项基本技巧，使我们能够对函数的乘积进行积分。它源自微分的乘积法则，通过策略性地选择哪一部分求导、哪一部分积分，将一个困难的积分转化为一个较简单的积分。在IB数学：分析与方法HL课程中，掌握这一方法对于处理从多项式乘指数函数到三角函数的循环积分等各种积分至关重要。

1. The Integration by Parts Formula | 分部积分公式

The integration by parts formula arises from the product rule for differentiation: (uv)’ = u’v + uv’. Integrating both sides and rearranging gives the standard form:

分部积分公式源于微分的乘积法则：(uv)’ = u’v + uv’。两边积分并整理后得到标准形式：

∫ u dv = uv − ∫ v du

Here, u and v are functions of x. The aim is to express a given integral in the form ∫ u dv, where u is chosen to simplify upon differentiation, and dv is chosen so that its integral v is easy to find. This formula transforms the product integral into a new integral that is hopefully simpler, plus a boundary term uv.

这里 u 和 v 是 x 的函数。目的是将给定的积分表示为 ∫ u dv 的形式，其中选择的 u 求导后会简化，而选择的 dv 积分后 v 容易求得。该公式把乘积的积分转化为一个希望更简单的新积分，再加上一个边界项 uv。

2. Choosing u and dv: The LIATE Rule | 选择u和dv：LIATE法则

A key strategic decision is selecting u and dv. The LIATE rule provides a helpful order of priority for choosing u: Logarithmic functions, Inverse trigonometric functions, Algebraic functions, Trigonometric functions, Exponential functions. The function appearing first in this list is typically chosen as u.

关键的战略决策是选择 u 和 dv。LIATE法则提供了一个选择 u 的有用优先级顺序：对数函数、反三角函数、代数函数、三角函数、指数函数。在该列表中排在前面的函数通常被选作 u。

For example, in ∫ x² ln x dx, ln x is logarithmic, x² is algebraic; since L comes before A, choose u = ln x. In ∫ eˣ sin x dx, trigonometric (T) appears before exponential (E), so u = sin x would be the conventional choice, though the integral cycles regardless if applied consistently.

例如，在 ∫ x² ln x dx 中，ln x 是对数函数，x² 是代数函数；因为 L 在 A 之前，所以选择 u = ln x。在 ∫ eˣ sin x dx 中，三角函数（T）出现在指数函数（E）之前，因此常规选择 u = sin x，不过只要应用一致，该积分会循环。

If an inverse trigonometric function is present, such as arctan x, it almost always becomes u because I comes very early. The LIATE rule is not absolute but is an excellent starting guide.

如果存在反三角函数，如 arctan x，它几乎总是被选为 u，因为 I 排位很靠前。LIATE法则并非绝对，但它是极佳的起始指导。

3. Basic Examples: Polynomial × Exponential/Trig | 基本例题：多项式乘指数/三角函数

When integrating x eˣ or x sin x, the algebraic part x is chosen as u (since A before T/E), and dv is eˣ dx or sin x dx. This reduces the polynomial degree in the new integral.

积分 x eˣ 或 x sin x 时，代数部分 x 选作 u（因为A在T/E之前），dv 为 eˣ dx 或 sin x dx。这样会降低新积分中多项式的次数。

Example: Evaluate ∫ x eˣ dx.
Solution: Let u = x, dv = eˣ dx. Then du = dx, v = eˣ.
Using the formula: ∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C.

例题：计算 ∫ x eˣ dx。
解：令 u = x, dv = eˣ dx。则 du = dx, v = eˣ。
代入公式：∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C。

For ∫ x cos x dx, set u = x, dv = cos x dx. Then v = sin x. So the integral becomes x sin x − ∫ sin x dx = x sin x + cos x + C.

对于 ∫ x cos x dx，设 u = x, dv = cos x dx。则 v = sin x。因此积分变为 x sin x − ∫ sin x dx = x sin x + cos x + C。

4. Integrating Logarithmic Functions | 积分对数函数

The natural logarithm function ln x can be integrated by parts by writing it as 1 · ln x and choosing u = ln x, dv = dx. Then du = (1/x) dx, v = x. The resulting integral cancels the x term:

自然对数函数 ln x 可以通过把它写成 1·ln x，并选择 u = ln x, dv = dx 来分部积分。于是 du = (1/x) dx, v = x。得出的积分抵消了 x 项：

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

For higher powers, such as ∫ xⁿ ln x dx with n ≠ −1, set u = ln x, dv = xⁿ dx. Then v = xⁿ⁺¹/(n+1) and the integral reduces to a simple power integral.

对于更高次幂，例如 ∫ xⁿ ln x dx (n ≠ −1)，设 u = ln x, dv = xⁿ dx。则 v = xⁿ⁺¹/(n+1)，积分就化简为简单的幂积分。

Example: ∫ x² ln x dx → u = ln x, dv = x² dx → du = (1/x) dx, v = x³/3. Then ∫ x² ln x dx = (x³/3) ln x − ∫ (x³/3)(1/x) dx = (x³ ln x)/3 − (1/3) ∫ x² dx = (x³ ln x)/3 − x³/9 + C.

例题：∫ x² ln x dx → u = ln x, dv = x² dx → du = (1/x) dx, v = x³/3。则 ∫ x² ln x dx = (x³/3) ln x − ∫ (x

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