📚 Integration by Parts: Derivation and Applications | 分部积分法的推导与应用
Integration by parts is one of the most powerful techniques in IB Mathematics AA HL. It transforms a difficult integral into a simpler one by reversing the product rule for differentiation. This article will guide you through its derivation, the art of choosing u and dv, and its application to a variety of exam-style problems.
分部积分法是IB数学AA HL中最强大的技巧之一。它通过逆转微分中的乘积法则，将一个复杂的积分转化为更简单的积分。本文将引导你完成它的推导、选择 u 与 dv 的技巧，以及它在各类考试题型中的应用。
1. Derivation from the Product Rule | 从乘积法则推导
Recall the product rule for differentiation: (uv)' = u'v + uv', where u and v are differentiable functions of x. Integrating both sides with respect to x gives ∫ (uv)' dx = ∫ u'v dx + ∫ uv' dx. The left side is simply uv. Rearranging yields the integration by parts formula.
回顾微积分中的乘积法则：(uv)' = u'v + uv'，其中 u 和 v 是关于 x 的可导函数。两边同时对 x 积分，得到 ∫ (uv)' dx = ∫ u'v dx + ∫ uv' dx。左边就是 uv。整理后即可得到分部积分公式。
∫ u dv = uv − ∫ v du
The formula is usually written in two equivalent forms. The first uses differentials, and the second uses derivatives: ∫ u(x)v'(x) dx = u(x)v(x) - ∫ v(x)u'(x) dx. Both appear in IB exams and formula booklets.
该公式通常有两种等价写法。第一种使用微分，第二种使用导数：∫ u(x)v'(x) dx = u(x)v(x) - ∫ v(x)u'(x) dx。这两种形式都出现在IB考试和公式手册中。
2. Understanding the Formula | 理解公式的结构
The key idea is to split the integrand into two parts: one that we differentiate (u) and one that we integrate (dv). The original integral ∫ u dv is replaced by the product uv minus a new integral ∫ v du. The goal is to make the new integral easier than the original.
核心思想是将被积函数拆成两部分：一部分用于求导（u），另一部分用于积分（dv）。原来的积分 ∫ u dv 被替换为乘积 uv 减去一个新的积分 ∫ v du。目标是使新的积分比原来的积分更简单。
Notice that du = u' dx and v = ∫ dv. Therefore, the choice of u and dv completely determines the transformed integral. A poor choice can make the new integral even more complicated.
注意 du = u' dx，且 v = ∫ dv。因此，u 和 dv 的选择完全决定了变换后的积分。选择不当会让新积分变得更加复杂。
3. Choosing u and dv | 如何选择 u 与 dv
A common mnemonic for selecting u is LIATE, which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential. Functions appearing earlier in the list are better choices for u, because their derivatives become simpler or more manageable.
选择 u 的常用口诀是LIATE，代表对数函数、反三角函数、代数函数、三角函数和指数函数。列表中越靠前的函数越适合作为 u，因为它们的导数会变得更简单或更易处理。
- L — Logarithmic functions: ln x, loga x
- I — Inverse trigonometric functions: arcsin x, arctan x
- A — Algebraic functions: xn, x2+1
- T — Trigonometric functions: sin x, cos x
- E — Exponential functions: ex, 2x
For example, in ∫ x ex dx, x is algebraic and ex is exponential. Since A comes before E in LIATE, choose u = x and dv = ex dx. Then du = dx and v = ex. The new integral ∫ ex dx is trivial.
例如，在 ∫ x ex dx 中，x 是代数函数，而 ex 是指数函数。由于在LIATE中A排在E之前，因此选择 u = x，dv = ex dx。于是 du = dx，v = ex。新的积分 ∫ ex dx 就非常简单了。
4. Classic Example: Polynomial × Exponential | 经典例题：多项式乘以指数函数
Let us compute ∫ x ex dx step by step. Following LIATE, set u = x and dv = ex dx. Then du = dx and v = ex. Substituting into the formula gives:
我们来逐步计算 ∫ x ex dx。根据LIATE，令 u = x，dv = ex dx。则 du = dx，v = ex。代入公式得到：
∫ x ex dx = x ex − ∫ ex dx = x ex − ex + C
This example shows the essence of integration by parts: the degree of the polynomial decreases from 1 to 0 after one application. For higher-degree polynomials, repeated application is needed.
这个例子展示了分部积分法的本质：多项式的次数在一次应用后从1降为0。对于更高次的多项式，需要重复应用。
5. Polynomial × Trigonometric Functions | 多项式乘以三角函数
Consider ∫ x cos x dx. Here x is algebraic and cos x is trigonometric. LIATE suggests u = x and dv = cos x dx. Thus du = dx and v = sin x. Applying the formula:
考虑 ∫ x cos x dx。这里 x 是代数函数，cos x 是三角函数。LIATE建议取 u = x，dv = cos x dx。于是 du = dx，v = sin x。代入公式：
∫ x cos x dx = x sin x − ∫ sin x dx = x sin x + cos x + C
Notice that the final result contains both sin x and cos x. This is typical when integrating products of polynomials and trigonometric functions.
注意最终结果同时包含 sin x 和 cos x。这是积分多项式与三角函数乘积时的典型特征。
6. Logarithmic and Inverse Trigonometric Functions | 对数和反三角函数
Integration by parts is essential for integrals such as ∫ ln x dx and ∫ arctan x dx. In these cases, we treat the integrand as 1 × f(x), choosing u = ln x or u = arctan x, and dv = dx.
分部积分法对于 ∫ ln x dx 和 ∫ arctan x dx 这类积分至关重要。在这些情况下，我们将被积函数视为 1 × f(x)，选择 u = ln x 或 u = arctan x，以及 dv = dx。
For ∫ ln x dx, set u = ln x and dv = dx. Then du = 1x dx and v = x. The formula yields:
对于 ∫ ln x dx，令 u = ln x，dv = dx。则 du = 1x dx，v = x。公式给出：
∫ ln x dx = x ln x − ∫ x · (1/x) dx = x ln x − x + C
Similarly, ∫ arctan x dx requires u = arctan x, dv = dx. This leads to an integral of a rational function that can be solved by substitution or partial fractions.
类似地，∫ arctan x dx 需要取 u = arctan x，dv = dx。这会得到一个有理函数的积分，可以通过换元法或部分分式法求解。
7. Repeated Application | 重复应用分部积分
When u is a polynomial of degree 2 or higher, one application of integration by parts may not be enough. For example, ∫ x2 ex dx requires two applications. First set u = x2, dv = ex dx. Then:
当 u 是2次或更高次的多项式时，一次分部积分可能不够。例如 ∫ x2 ex dx 需要应用两次。首先令 u = x2，dv = ex dx。则：
∫ x² eˣ dx = x² eˣ − 2 ∫ x eˣ dx
We already know ∫ x ex dx = (x - 1)ex + C. Substituting this back gives:
我们已经知道 ∫ x ex dx = (x - 1)ex + C。代回后得到：
∫ x² eˣ dx = x² eˣ − 2(x − 1)eˣ + C = (x² − 2x + 2)eˣ + C
In IB exams, you may be asked to apply the technique twice or even three times. Keep track of signs carefully, especially when integrating trigonometric functions.
在IB考试中，你可能会被要求应用两次甚至三次。要仔细注意符号，特别是在积分三角函数时。
8. Cycling Integrals | 循环积分
Some integrals lead back to the original integral after two applications. A classic example is I = ∫ ex sin x dx. Both functions are transcendental, so neither choice is obviously better. Choose u = ex, dv = sin x dx. Then:
有些积分在应用两次后会回到原始积分。经典例子是 I = ∫ ex sin x dx。两个函数都是超越函数，因此选择没有明显优劣。取 u = ex，dv = sin x dx。则：
I = −eˣ cos x + ∫ eˣ cos x dx
Apply integration by parts again to the new integral with u = ex, dv = cos x dx:
对新的积分再次应用分部积分，令 u = ex，dv = cos x dx：
∫ eˣ cos x dx = eˣ sin x − ∫ eˣ sin x dx = eˣ sin x − I
Substituting back gives I = -ex cos x + ex sin x - I. Solve for I:
代回得到 I = -ex cos x + ex sin x - I。解出 I：
2I = eˣ(sin x − cos x) + C
∫ eˣ sin x dx = ½ eˣ(sin x − cos x) + C
This "cyclical" method is a common IB question type and also appears in paper 2 calculators often.
这种“循环”方法是IB常见的题型类型，也常出现在卷2（可使用计算器）中。
9. Definite Integrals and Evaluation | 定积分与求值
For definite integrals, the formula becomes ∫ab u dv = [uv]ab - ∫ab v du. The boundary term [uv]ab must be evaluated carefully.
对于定积分，公式变为 ∫ab u dv = [uv]ab - ∫ab v du。边界项 [uv]ab 必须仔细计算。
Example: ∫01 x ex dx. We already know the antiderivative is (x - 1)ex. Evaluate:
例：∫01 x ex dx。我们已知原函数是 (x - 1)ex。计算：
[(x − 1)eˣ]₀¹ = (0) · e¹ − (−1) · e⁰ = 1
In IB exams, definite integrals often appear in questions about areas under curves, volumes of revolution, or probability distributions. Always remember to include the limits when applying integration by parts.
在IB考试中，定积分常出现在曲线下面积、旋转体体积或概率分布问题中。应用分部积分时，务必带上积分上下限。
10. Combining with Substitution | 与换元法结合
Sometimes integration by parts must be preceded or followed by a substitution. For example, ∫ arcsin x dx can be simplified by choosing u = arcsin x, which leads to ∫ x⁄√1-x2 dx. This remaining integral is solved by the substitution t = 1 - x2, giving -√1-x2.
有时分部积分需要与换元法结合使用。例如 ∫ arcsin x dx 可以选 u = arcsin x，从而得到 ∫ x⁄√1-x2 dx。剩余的积分用换元 t = 1 - x2 来解，结果为 -√1-x2。
Another common combination is ∫ x ln(x+1) dx. Choose u = ln(x+1), dv = x dx. Then du = 1x+1 dx and v = x2⁄2. The resulting integral may require polynomial division or substitution.
另一个常见组合是 ∫ x ln(x+1) dx。取 u = ln(x+1)，dv = x dx。则 du = 1x+1 dx，v = x2⁄2。得到的积分可能需要多项式除法或换元法。
11. Common Mistakes and Exam Tips | 常见错误与考试技巧
A frequent error is choosing the wrong u, which leads to a more complicated integral. Always test whether the derivative of u is simpler than u itself. Another common mistake is forgetting the negative sign in front of the new integral, especially when integrating trigonometric functions.
一个常见错误是选择了错误的 u，导致积分变得更复杂。始终检查 u 的导数是否比 u 本身更简单。另一个常见错误是忘记新积分前面的负号，特别是在积分三角函数时。
- Simplify before integrating: Rewrite ln x⁄x2 as ln x · x-2 to clearly identify u = ln x.
- Check by differentiating: After finding the antiderivative, differentiate it to verify the result.
- Be careful with signs: Write down every step to avoid losing minus signs.
- Use LIATE consistently: It is not a law, but it works for most IB questions.
在积分前先化简：将 ln x⁄x2 写成 ln x · x-2，以便清楚识别 u = ln x。通过求导检验结果：得到原函数后，对其求导验证是否正确。注意符号：写出每一步以避免漏掉负号。一致使用LIATE：它并非法则，但对大多数IB题目有效。
12. Summary and Final Words | 总结与结语
Integration by parts is a fundamental technique that every IB Mathematics AA HL student must master. Its derivation from the product rule makes it logical and easy to remember. The key is to choose u and dv wisely, usually following the LIATE rule, and to practise repetitive and cyclical cases until they become automatic.
分部积分法是每位IB数学AA HL学生都必须掌握的基本技巧。其推导来源于乘积法则，因此逻辑清晰且易于记忆。关键在于明智地选择 u 与 dv，通常遵循LIATE法则，并通过反复练习循环与重复情形，直至熟练自如。
In exams, show all steps clearly, especially the substitution into the formula. A well-structured solution not only earns full marks but also helps you avoid sign errors. With consistent practice, integration by parts will become a reliable tool in your mathematical toolbox.
考试中要清晰地展示所有步骤，特别是代入公式的过程。结构清晰的解答不仅能获得满分，还能帮助你避免符号错误。通过持续练习，分部积分法将成为你数学工具箱中可靠的工具。
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