Integration by Parts | 分部积分法

📚 Integration by Parts | 分部积分法

Integration by parts is a fundamental technique for integrating products of functions. It transforms a difficult integral into a simpler one, based on the product rule for differentiation. In IB Mathematics: Analysis and Approaches (Higher Level), you will routinely apply this method to integrands such as x sin x, e^x cos x, and ln x. Mastering the choice of u and dv is the key to efficient and accurate computation.

分部积分法是处理函数乘积积分的基本技巧。它将复杂的积分转化为更简单的积分，其原理源于微分的乘积法则。在 IB 数学：分析与方法（高水平）课程中，你将经常用此方法处理诸如 x sin x、e^x cos x 和 ln x 等被积函数。熟练掌握 u 与 dv 的选择是实现高效准确计算的关键。

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1. The Product Rule in Reverse | 逆向乘积法则

The product rule states that d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x). Integrating both sides with respect to x yields ∫ u'(x)v(x) dx + ∫ u(x)v'(x) dx = u(x)v(x) + C. Rearranging, we obtain ∫ u(x)v'(x) dx = u(x)v(x) − ∫ u'(x)v(x) dx. Letting u = u(x) and dv = v'(x) dx gives du = u'(x) dx and v = v(x). This produces the compact formula:

乘积法则指出 d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)。两边同时对 x 积分，得 ∫ u'(x)v(x) dx + ∫ u(x)v'(x) dx = u(x)v(x) + C。整理后得到 ∫ u(x)v'(x) dx = u(x)v(x) − ∫ u'(x)v(x) dx。令 u = u(x)，dv = v'(x) dx，便有 du = u'(x) dx 以及 v = v(x)。由此便得到了简洁的公式：

∫ u dv = uv − ∫ v du

In IB examinations, you will often see it written as ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx. The art of using integration by parts lies in splitting the original integrand so that the new integral ∫ v du is easier to evaluate than the original one.

在 IB 考试中，你常会见到形式 ∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx。使用分部积分法的技巧在于合理拆分原被积函数，使得新积分 ∫ v du 比原积分更容易计算。

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2. The LIATE Rule for Choosing u | 选择 u 的 LIATE 法则

Choosing which function to set as u can be simplified by the LIATE order of precedence. This acronym helps you decide which type of function typically yields a simpler derivative and thus a more manageable new integral. The rule is:

选择哪个函数作为 u 可以通过 LIATE 优先顺序来简化。这个缩写帮助你判断哪种函数类型通常能得到更简单的导数，从而产生更易处理的新积分。其顺序为：

Letter	Function type (English)	函数类型（中文）
L	Logarithmic (ln x, logax)	对数函数
I	Inverse trigonometric (arcsin x, arctan x)	反三角函数
A	Algebraic (xn, polynomials)	代数函数
T	Trigonometric (sin x, cos x)	三角函数
E	Exponential (ex, ax)	指数函数

Generally, if you have a product of two functions from different categories, choose u as the one that appears earlier in LIATE. For example, in ∫ x² ln x dx, ln x is Logarithmic (L) and x² is Algebraic (A), so select u = ln x and dv = x² dx. This rule covers most IB cases, though some cyclic integrals (with e^x and sin x) are more flexible.

通常，若被积函数属于两个不同类别的函数相乘，应将 LIATE 中靠前的函数选作 u。例如，在 ∫ x² ln x dx 中，ln x 是对数函数（L），x² 是代数函数（A），因此应选 u = ln x，dv = x² dx。此规则涵盖了大多数 IB 考题，但对于一些循环积分（如 e^x 与 sin x 的乘积），选择更灵活。

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3. First Example: ∫ x e^x dx | 第一个例题：∫ x e^x dx

This is a classic introduction to integration by parts. The integrand is the product of an algebraic function x and an exponential function e^x. According to LIATE, Algebraic comes before Exponential, so we set u = x and dv = e^x dx.

这是分部积分法的经典入门例题。被积函数是代数函数 x 与指数函数 e^x 的乘积。根据 LIATE 法则，代数函数优先于指数函数，因此我们令 u = x，dv = e^x dx。

Then du = dx, and integrating dv gives v = e^x. Substituting into the formula ∫ u dv = uv − ∫ v du yields:

则 du = dx，对 dv 积分得 v = e^x。代入公式 ∫ u dv = uv − ∫ v du 得到：

∫ x e^x dx = x e^x − ∫ e^x dx = x e^x − e^x + C

Notice how the new integral ∫ e^x dx is far simpler than the original. Factoring out e^x gives the neat result e^x(x − 1) + C. Always check your answer by differentiating: the derivative of e^x(x − 1) is e^x(x − 1) + e^x = x e^x, which confirms the solution.

注意新积分 ∫ e^x dx 比原积分简单得多。将 e^x 提取公因子可得简洁结果 e^x(x − 1) + C。务必通过求导检查答案：e^x(x − 1) 的导数为 e^x(x − 1) + e^x = x e^x，验证了结果的正确性。

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4. Integrating ln x and Logarithmic Functions | 积分 ln x 与对数函数

The integral ∫ ln x dx does not appear to be a product, but we can treat it as ∫ (1)·ln x dx. Here, u = ln x (L in LIATE) and dv = dx, so v = x and du = (1/x) dx.

积分 ∫ ln x dx 看起来不像乘积，但可将其视为 ∫ (1)·ln x dx。这里 u = ln x（LIATE 中的 L），dv = dx，因此 v = x，du = (1/x) dx。

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

This result is worth memorising. For definite integrals, such as ∫₁^e ln x dx, apply the antiderivative and evaluate: [x ln x − x]₁^e = (e·1 − e) − (1·0 − 1) = 0 − (−1) = 1. The same technique works for ∫ ln(ax + b) dx, though a substitution may be needed first.

这个结果值得记忆。对于定积分，如 ∫₁^e ln x dx，应用原函数并求值：[x ln x − x]₁^e = (e·1 − e) − (1·0 − 1) = 0 − (−1) = 1。同样的技巧适用于 ∫ ln(ax + b) dx，但可能需要先进行换元。

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5. Integrating Inverse Trigonometric Functions | 积分反三角函数

Integrals such as ∫ arctan x dx and ∫ arcsin x dx are handled similarly to ln x. For arctan x, set u = arctan x and dv = dx. Then du = 1/(1 + x²) dx and v = x.

∫ arctan x dx 和 ∫ arcsin x dx 这类积分的处理方法与 ln x 类似。对于 arctan x，令 u = arctan x，dv = dx。则 du = 1/(1 + x²) dx，v = x。

∫ arctan x dx = x arctan x − ∫ x/(1 + x²) dx

The remaining integral can be evaluated by recognising that the derivative of 1 + x² is 2x, so ∫ x/(1 + x²) dx = ½ ln(1 + x²) + C. Hence, ∫ arctan x dx = x arctan x − ½ ln(1 + x²) + C. For arcsin x, a similar approach yields x arcsin x + √(1 − x²) + C, which also appears in IB formula booklets.

剩下的积分可这样计算：因 1 + x² 的导数为 2x，故 ∫ x/(1 + x²) dx = ½ ln(1 + x²)

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