📚 Mastering Integration by Parts for IB Math | IB数学：分部积分法解题技巧
Integration by parts is a powerful technique that stems from the product rule in differentiation. It allows us to integrate products of functions and even single logarithmic or inverse trigonometric functions that no basic method can handle. Mastering this tool is essential for IB Math AA HL (and SL for some functions), as it frequently appears in exams, particularly in Paper 2 and Paper 3.
分部积分法是微积分中一种强大的工具，它来源于微分中的乘法法则。它使我们能够对两个函数的乘积进行积分，甚至可以处理单独的对数函数或反三角函数，这些是无法通过基础方法解决的。掌握这一技巧对于 IB 数学分析与方法 HL（以及部分 SL）至关重要，因为它经常出现在考试中，特别是在 Paper 2 和 Paper 3 中。
1. The Core Formula | 核心公式与选择标准
The standard formula is derived from the product rule. If you have two differentiable functions, the integration by parts formula states: the integral of u dv equals u v minus the integral of v du. The key lies in correctly identifying which part of the integrand should be ‘u’ (which you will differentiate) and which should be ‘dv’ (which you will integrate).
标准公式来源于乘法法则。如果你有两个可微函数，分部积分法公式为：u 对 v 的积分等于 u 乘以 v 减去 v 对 u 的积分。关键在于正确识别被积函数中哪一部分应作为 ‘u’（你将对其求导），哪一部分应作为 ‘dv’（你将对其进行积分）。
∫ u dv = uv – ∫ v du
2. The LIATE Rule | LIATE 选择原则
LIATE is a helpful acronym for deciding which function to choose as ‘u’. The priority for ‘u’ follows this order: Logarithmic, Inverse trigonometric, Algebraic (polynomials), Trigonometric, and Exponential. Choosing ‘u’ in this order ensures the new integral ∫ v du is generally simpler than the original.
LIATE 是一个选择哪个函数作为 ‘u’ 的实用缩略词。’u’ 的优先级顺序为：Logarithmic 对数、Inverse trigonometric 反三角、Algebraic 代数（多项式）、Trigonometric 三角、Exponential 指数。按此顺序选择 ‘u’ 可以确保新的积分 ∫ v du 通常比原积分更简单。
- L: Logarithmic functions, e.g., ln x, logₐ x | 对数函数，例如 ln x， logₐ x
- I: Inverse trigonometric functions, e.g., arctan x, arcsin x | 反三角函数，例如 arctan x， arcsin x
- A: Algebraic functions, e.g., x², x⁵ | 代数函数，例如 x²， x⁵
- T: Trigonometric functions, e.g., sin x, cos x | 三角函数，例如 sin x， cos x
- E: Exponential functions, e.g., eˣ, 2ˣ | 指数函数，例如 eˣ， 2ˣ
3. Algebraic × Exponential | 基础代数函数乘以指数函数
This is the most straightforward application of integration by parts. Consider the integral of x times e to the x. According to LIATE, ‘x’ is Algebraic and should be chosen as ‘u’. Let u equal x and dv equal e to the x dx. Then du equals dx and v equals e to the x.
这是分部积分法最直接的应用。考虑 x 乘以 e 的 x 次方的积分。根据 LIATE 原则，’x’ 是代数函数，应作为 ‘u’。令 u 等于 x，dv 等于 e 的 x 次方 dx。则 du 等于 dx，v 等于 e 的 x 次方。
∫ x eˣ dx = x eˣ – ∫ eˣ dx = x eˣ – eˣ + C
Notice how the power of x decreased from one to zero, making the remaining integral trivial. This demonstrates the power of choosing the algebraic term as ‘u’.
注意到 x 的幂次从 1 降到了 0，使得剩下的积分变得非常简单。这体现了选择代数项作为 ‘u’ 的优势。
4. Polynomial × Logarithm | 代数函数乘以对数函数
Logarithms are the top priority in the LIATE rule. For the integral of x squared times ln x, you must set u equal to ln x and dv equal to x squared dx. This is because differentiating ln x gives a simple reciprocal, while integrating x squared is straightforward.
对数函数在 LIATE 原则中优先级最高。对于 x 的平方乘以 ln x 的积分，你必须设 u 等于 ln x，dv 等于 x 的平方 dx。这是因为对 ln x 求导得到简单的倒数，而对 x 的平方积分也很直接。
∫ x² ln x dx = (x³/3) ln x – ∫ (x³/3)(1/x) dx = (x³/3) ln x – x³/9 + C
If you chose u equal to x squared instead, the integral would become more complicated. Therefore, practicing the LIATE classification is an invaluable skill for solving IB questions quickly.
如果你错误地选择 u 等于 x 的平方，积分会变得更加复杂。因此，练习 LIATE 分类法对于在 IB 考试中快速解题是一项非常有价值的技能。
5. Single Logarithm or Inverse Trig | 单一对数或反三角函数
When you only have one function that isn’t obviously integrable into a standard form, you can treat it as itself multiplied by 1. For the integral of ln x dx, set u equal to ln x and dv equal to 1 dx. This technique effectively uses the constant 1 as the second function.
当你只有一个不能直接积分的函数时，你可以将其视为它本身乘以 1。对于 ln x dx 的积分，令 u 等于 ln x，dv 等于 1 dx。这个技巧有效地将常数 1 作为第二个函数。
∫ ln x dx = x ln x – ∫ x · (1/x) dx = x ln x – x + C
Similarly, for the integral of arctan x dx, set u equal to arctan x and dv equal to 1 dx. Since du equals 1/(1+x²) dx and v equals x, the resulting integral ∫ x/(1+x²) dx can be solved by a simple substitution.
类似地，对于 arctan x dx 的积分，令 u 等于 arctan x，dv 等于 1 dx。因为 du 等于 1/(1+x²) dx，v 等于 x，所以得到的积分 ∫ x/(1+x²) dx 可以通过简单的换元法求解。
6. Circular Integration | 循环积分法
Sometimes, applying integration by parts twice leads you back to the original integral. This often happens with exponential and trigonometric functions. Consider the integral of e to the x times sin x dx.
有时，两次应用分部积分法会让你回到原积分。这经常发生在指数函数和三角函数的乘积中。考虑 e 的 x 次方乘以 sin x 的积分。
I = ∫ eˣ sin x dx
Let u equal sin x and dv equal eˣ dx. Then du equals cos x dx and v equals eˣ. Applying the formula gives I equals eˣ sin x minus the integral of eˣ cos x dx. Let J represent this new integral. Applying integration by parts again to J, we find that J equals eˣ cos x plus the original integral I.
令 u 等于 sin x，dv 等于 eˣ dx。则 du 等于 cos x dx，v 等于 eˣ。代入公式得到 I 等于 eˣ sin x 减去 eˣ cos x dx 的积分。令 J 代表这个新的积分。再次对 J 应用分部积分法，我们发现 J 等于 eˣ cos x 加上原积分 I。
I = eˣ sin x – eˣ cos x – I → 2I = eˣ (sin x – cos x)
Solving algebraically yields the final answer: I equals one half e to the x times (sin x minus cos x), plus the constant of integration. This algebraic manipulation is a classic IB exam question.
通过代数运算求解得到最终答案：I 等于二分之一 e 的 x 次方乘以 (sin x 减 cos x)，再加上积分常数。这种代数操作是一个经典的 IB 考试题型。
7. Definite Integrals | 定积分的处理
For definite integrals, you must apply the limits to the ‘uv’ part and to the resulting integral ∫ v du. A common trick to stay organized is to create a small table of u, du, v, and dv before substituting. This minimizes careless mistakes under exam pressure.
对于定积分，你必须将上下限同时代入 ‘uv’ 部分以及新积分 ∫ v du。一个保持条理清晰的常见技巧是在代入前建立一个小表格，列出 u、du、v 和 dv。这能最大程度地减少考试压力下的粗心错误。
Consider the integral from 0 to 1 of x e to the x dx. First find the indefinite integral, x e to the x minus e to the x, then evaluate it at the upper and lower limits.
考虑从 0 到 1 的 x e 的 x 次方 dx 的积分。首先求不定积分 x e 的 x 次方减去 e 的 x 次方，然后在上下限处计算差值。
∫₀¹ x eˣ dx = [x eˣ – eˣ]₀¹ = (1·e – e) – (0·1 – 1) = 0 – (-1) = 1
8. Recursion Formulas | 递推公式的推导
IB Higher Level exams often ask for patterns. For a sequence of integrals I sub n equals the integral of x to the n times e to the x dx, you can use integration by parts to derive a reduction formula.
IB 高级水平考试经常考察规律。对于积分序列 Iₙ 等于 x 的 n 次方乘以 e 的 x 次方 dx 的积分，你可以使用分部积分法推导出递推公式。
Iₙ = ∫ xⁿ eˣ dx = xⁿ eˣ – n Iₙ₋₁
This is achieved by setting u equal to x to the n and dv equal to e to the x dx. The power of x decreases by one each time, allowing you to express a complex integral in terms of a simpler one. This type of proof is a common Paper 3 question.
这是通过令 u 等于 x 的 n 次方，dv 等于 e 的 x 次方 dx 实现的。x 的幂次每次都减少 1，从而可以用更简单的积分表示复杂的积分。这种证明是 Paper 3 的常见题型。
9. Combining with Substitution | 与换元法结合使用
Sometimes, an integral is best solved by substituting first, then integrating by parts to simplify the result. A classic example is the integral of sin of the square root of x dx. Initially, there is no clear product of functions, but a substitution reveals a hidden structure.
有时，先换元再分部积分会使积分更容易求解。一个经典的例子是 sin 根号 x 的积分。起初，没有明显的函数乘积，但换元揭示了隐藏的结构。
Let t = √x → x = t² → dx = 2t dt
∫ sin√x dx = 2 ∫ t sin t dt
Now, apply integration by parts to the new integral. Let u equal t and dv equal sin t dt. The final result is 2 sin√x minus 2√x cos√x, plus C. Recognizing when to combine techniques is a key higher-level skill.
现在，对新积分应用分部积分法。令 u 等于 t，dv 等于 sin t dt。最终结果是 2 sin√x 减去 2√x cos√x，再加上 C。识别何时结合不同的技巧是高级水平的关键能力。
10. Common Pitfalls | 常见错误与陷阱
There are several common mistakes that students make when applying integration by parts. Being aware of these can save you valuable marks. First, choosing the wrong ‘u’ can increase the power of x instead of decreasing it, leading to a more complex integral.
学生在应用分部积分法时经常会犯几个常见错误。意识到这些错误可以帮你节省宝贵的
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