Integration by Parts: Derivation and Applications | 分部积分法的推导与应用

📚 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 \(\int (uv)’ \, dx = \int u’v \, dx + \int uv’ \, dx\). 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 \(\int (uv)’ \, dx = \int u’v \, dx + \int uv’ \, dx\)。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: \(\int u(x)v'(x) \, dx = u(x)v(x) – \int v(x)u'(x) \, dx\). 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：\(\int u(x)v'(x) \, dx = u(x)v(x) – \int v(x)u'(x) \, dx\)。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 \(\int u \, dv\) is replaced by the product \(uv\) minus a new integral \(\int v \, du\). 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 \(\int u \, dv\) 被替换为乘积 \(uv\) 减去一个新的积分 \(\int v \, du\)。目标是使新的积分比原来的积分更简单。

Notice that \(du = u’ \, dx\) and \(v = \int dv\). 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 \(v = \int dv\)。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: \(\ln x, \log_a x\)
I — Inverse trigonometric functions: \(\arcsin x, \arctan x\)
A — Algebraic functions: \(x^n, x^2+1\)
T — Trigonometric functions: \(\sin x, \cos x\)
E — Exponential functions: \(e^x, 2^x\)

For example, in \(\int x e^x \, dx\), \(x\) is algebraic and \(e^x\) is exponential. Since A comes before E in LIATE, choose \(u = x\) and \(dv = e^x \, dx\). Then \(du = dx\) and \(v = e^x\). The new integral \(\int e^x \, dx\) is trivial.

例如，在 \(\int x e^x \, dx\) 中，\(x\) 是代数函数，而 \(e^x\) 是指数函数。由于在LIATE中A排在E之前，因此选择 \(u = x\)，\(dv = e^x \, dx\)。于是 \(du = dx\)，\(v = e^x\)。新的积分 \(\int e^x \, dx\) 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 \(\int x e^x \, dx\) step by step. Following LIATE, set \(u = x\) and \(dv = e^x \, dx\). Then \(du = dx\) and \(v = e^x\). Substituting into the formula gives:

我们来逐步计算 \(\int x e^x \, dx\)。根据LIATE，令 \(u = x\)，\(dv = e^x \, dx\)。则 \(du = dx\)，\(v = e^x\)。代入公式得到：

∫ x e^x dx = x e^x − ∫ e^x dx = x e^x − e^x + 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 \(\int x \cos x \, dx\). 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 \(\int x \cos x \, dx\)。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 \(\int \ln x \, dx\) and \(\int \arctan x \, dx\). In these cases, we treat the integrand as \(1 \times f(x)\), choosing \(u = \ln x\) or \(u = \arctan x\), and \(dv = dx\).

分部积分法对于 \(\int \ln x \, dx\) 和 \(\int \arctan x \, dx\) 这类积分至关重要。在这些情况下，我们将被积函数视为 \(1 \times f(x)\)，选择 \(u = \ln x\) 或 \(u = \arctan x\)，以及 \(dv = dx\)。

For \(\int \ln x \, dx\), set \(u = \ln x\) and \(dv = dx\). Then \(du = \frac{1}{x} dx\) and \(v = x\). The formula yields:

对于 \(\int \ln x \, dx\)，令 \(u = \ln x\)，\(dv = dx\)。则 \(du = \frac{1}{x} dx\)，\(v = x\)。公式给出：

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

Similarly, \(\int \arctan x \, dx\) 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，\(\int \arctan x \, dx\) 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, \(\int x^2 e^x \, dx\) requires two applications. First set \(u = x^2\), \(dv = e^x \, dx\). 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 \(\int x^2 e^x \, dx\) 需要应用两次。首先令 \(u = x^2\)，\(dv = e^x \, dx\)。则：

∫ x² eˣ dx = x² eˣ − 2 ∫ x eˣ dx

We already know \(\int x e^x \, dx = (x – 1)e^x + C\). Substituting this back gives:

我们已经知道 \(\int x e^x \, dx = (x – 1)e^x + C\)。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 \(I = \int e^x \sin x \, dx\). Both functions are transcendental, so neither choice is obviously better. Choose \(u = e^x\), \(dv = \sin x \, dx\). Then:

有些积分在应用两次后会回到原始积分。经典例子是 \(I = \int e^x \sin x \, dx\)。两个函数都是超越函数，因此选择没有明显优劣。取 \(u = e^x\)，\(dv = \sin x \, dx\)。则：

I = −eˣ cos x + ∫ eˣ cos x dx

Apply integration by parts again to the new integral with \(u = e^x\), \(dv = \cos x \, dx\):

对新的积分再次应用分部积分，令 \(u = e^x\)，\(dv = \cos x \, dx\)：

∫ eˣ cos x dx = eˣ sin x − ∫ eˣ sin x dx = eˣ sin x − I

Substituting back gives \(I = -e^x \cos x + e^x \sin x – I\). Solve for \(I\):

代回得到 \(I = -e^x \cos x + e^x \sin x – I\)。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 \(\int_a^b u \, dv = [uv]_a^b – \int_a^b v \, du\). The boundary term \([uv]_a^b\) must be evaluated carefully.

对于定积分，公式变为 \(\int_a^b u \, dv = [uv]_a^b – \int_a^b v \, du\)。边界项 \([uv]_a^b\) 必须仔细计算。

Example: \(\int_0^1 x e^x \, dx\). We already know the antiderivative is \((x – 1)e^x\). Evaluate:

例：\(\int_0^1 x e^x \, dx\)。我们已知原函数是 \((x – 1)e^x\)。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, \(\int \arcsin x \, dx\) can be simplified by choosing \(u = \arcsin x\), which leads to \(\int \frac{x}{\sqrt{1-x^2}} \, dx\). This remaining integral is solved by the substitution \(t = 1 – x^2\), giving \(-\sqrt{1-x^2}\).

有时分部积分需要与换元法结合使用。例如 \(\int \arcsin x \, dx\) 可以选 \(u = \arcsin x\)，从而得到 \(\int \frac{x}{\sqrt{1-x^2}} \, dx\)。剩余的积分用换元 \(t = 1 – x^2\) 来解，结果为 \(-\sqrt{1-x^2}\)。

Another common combination is \(\int x \ln(x+1) \, dx\). Choose \(u = \ln(x+1)\), \(dv = x \, dx\). Then \(du = \frac{1}{x+1} dx\) and \(v = \frac{x^2}{2}\). The resulting integral may require polynomial division or substitution.

另一个常见组合是 \(\int x \ln(x+1) \, dx\)。取 \(u = \ln(x+1)\)，\(dv = x \, dx\)。则 \(du = \frac{1}{x+1} dx\)，\(v = \frac{x^2}{2}\)。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 \(\frac{\ln x}{x^2}\) as \(\ln x \cdot x^{-2}\) 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 \(\frac{\ln x}{x^2}\) 写成 \(\ln x \cdot x^{-2}\)，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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Integration by Parts: Derivation and Applications | 分部积分法的推导与应用