Mastering Integration Techniques: Substitution and Parts | 掌握积分技巧：换元法与分部积分法

📚 Mastering Integration Techniques: Substitution and Parts | 掌握积分技巧：换元法与分部积分法

by substitution and integration by parts. Inspired by typical problem sets such as those collated in pdfjoiner_(4)-093, this article unpacks these techniques step by step. We start from the standard integrals, move through the logic of reversing the chain rule, and then master the product‑ finite integrals. Whether you are consolidating your Year 12 knowledge or preparing for the final pure papers, a fluent command of these methods will transform your ability to tackle both routine and synoptic questions.

积分是A-Level数学的基石，而Edexcel考纲中两个最强大的工具就是换元积分法和分部积分法。本文从典型题目集（如pdfjoiner_(4)-093中的练习）出发，逐步拆解这些技巧。我们从标准积分公式起步，理解反链式法则的逻辑，再掌握与乘积法则对应的分部积分法。文中配有详细示例、选取u和dv的策略提示以及处理定积分的建议。无论你是在巩固12年级的内容还是备战最终的纯数试卷，熟练运用这些方法都将极大地提升你应对常规题和综合性问题的能力。

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1. Standard Integrals – Your Building Blocks | 标准积分：你的基石

our memory. For polynomial terms, ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, provided n ≠ −1. When n = −1, the integral is ln|x| = −cos x + C and ∫ cos x dx = sin x + C. The exponential function is its own integral: ∫ eˣ dx = eˣ . Knowing these instantly frees up mental energy for the more complex manipulations ahead.

在学习换元法和分部法之前，你必须对Edexcel公式表及记忆中那些标准积分烂熟于心。对于多项式项，∫ xⁿ dx = xⁿ⁺¹/(n+1) + C，其中n ≠ −1；当n = −1时，积分结果是 ln|x| + C。三角积分包括 ∫ sin x dx = −cos x + C 和 ∫ cos x dx = sin x + C。指数函数是其自身的积分：∫ eˣ dx = eˣ + C。瞬间想起这些公式能让你在后续更复杂的操作中腾出脑力。

Function	Integral
xⁿ (n≠−1)	xⁿ⁺¹/(n+1) + C
1/x	ln|x| + C
eˣ	eˣ + C
sin x	−cos x + C
cos x	sin x + C
sec² x	tan x + C

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2. The Reverse Chain Rule – Why Substitution Works | 反链式法则：换元法的原理

on by substitution is simply reversing this process. If you spot an integral of the form ∫ f(g(x))·g'(x) dx, you can let u = g(x) so that du = g' the standard forms you already know. The key is to identify a function and its derivative – or a constant multiple of its derivative – hidden inside the integrand.

当你对复合函数如 sin(2x) 求导时，需要乘以内层函数的导数，结果是 2cos(2x)。换元积分法正是这一过程的逆运算。如果你发现积分形如 ∫ f(g(x))·g'(x) dx，就可以令 u = g(x)，从而 du = g'(x) dx，整个积分就变成了 ∫ f(u) du。这把一个可能很凌乱的表达式转变为你已经掌握的标准形式。关键在于识别出被积函数中隐藏的一个函数及其导数——或者导数的常数倍。

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3. Integration by Substitution – The Method Step by Step | 换元积分法：逐步操作

u³ du, which is (1/4)u⁴ + C. Substituting back gives (1/4)(x²+1)⁴ + C. For more complicated cases where the derivative does not match exactly, you can still use substitution but you may need to manipulate constants. For instance, in ∫ x·√2x+1 dx, set u = 2x+1, then x = (u−1)/2 and dx = du/2. The integral becomes ∫ ((u−1)/2)·√u −1)u¹⁄² du, which simplifies to (1/4)∫ (u³⁄² − u¹⁄²) du. This is now a sum of power rules.

考虑 ∫ 2x·(x²+1)³ dx。这里的内层函数是 x²+1，其导数 2x 恰好就在前面。令 u = x²+1，则 du = 2x dx。积分变为 ∫ u³ du，结果是 (1/4)u⁴ + C。代回原变量得 (1/4)(x²+1)⁴ + C。对于导数不完全匹配的复杂情形，你仍然可以使用换元法，但可能需要处理常数。例如，∫ x·√2x+1 dx 中，设 u = 2x+1，则 x = (u−1)/2，dx = du/2。积分变成 ∫ ((u−1)/2)·√u · (du/2) = (1/4)∫ (u−1)u¹⁄² du，化简为 (1/4)∫ (u³⁄² − u¹⁄²) du。这现在就成了一组幂函数的求和。

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4. Definite Integrals and Changing Limits | 定积分与积分限的变换

you to change the limits of integration. Suppose you are evaluating ∫ from 0 to 1 of 4x·e^2x² from u=0 to u=2 of eᵘ du = [eᵘ] from 0 to 2 = e² − 1. You must never mix the old variable's limits with the new variable' ert back to x at the end, reducing algebraic errors and making the working more efficient.

对于定积分，换元时你必须更换积分上下限。假设你要计算 ∫ 从0到1 的 4x·e^2x² dx。令 u = 2x²。当 x = 0 时 u = 0；当 x = 1 时 u = 2。同时 du = 4x dx，因此积分变成 ∫ 从u=0到u=2 的 eᵘ du = [eᵘ] 从0到2 = e² − 1。你绝不能将原变量的积分限与新变量的被积函数混用。变换积分限能避免最后再换回x的步骤，减少代数错误，使解题过程更高效。

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5. Integration by Parts – The Product Rule Reversed | 分部积分法：乘积法则的逆运算

/dx(uv) = u dv/dx + v du/dx. Rearranging and integrating both sides yields the integration‑by‑ ferent types of function multiplied together, such as a polynomial times an exponential, a logarithm, or a trigonometric function. The art lies in choosing which factor to differentiate and which to integrate.

微分学中有乘积法则：d/dx(uv) = u dv/dx + v du/dx。重新整理并对两边积分就得到了分部积分公式：∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx。更简洁的形式是 ∫ u dv = uv − ∫ v du。当被积函数由两类不同的函数相乘构成时——例如多项式乘以指数函数、对数函数或三角函数——这一技巧就极为宝贵。其中的艺术在于选择哪一部分求导、哪一部分积分。

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6. How to Choose u and dv – The LIATE Rule | 如何选择u与dv：LIATE法则

ons, Algebraic (polynomials), Trigonometric functions, Exponentials. Whichever function appears earlier in the list should usually be chosen as u, because its derivative tends to simplify the remaining integral. For example, in ∫ x·ln x dx, the logarithm comes before the polynomial, so set u = ln x and dv = x dx. Then du = (1/x) dx and v = x²/2, giving ∫ x·ln x dx = (x²/2)ln x − ∫ (x²/2)(1/x) dx = (x²/2)ln x − ∫ x/2 dx = (x²/2)ln x − x²/4 + C. In ∫ x·eˣ dx, however, the algebraic part x comes before the exponential, so u = x, dv = eˣ dx, and you get x·eˣ − ∫ eˣ dx = x·eˣ − eˣ + C.

Edexcel考生中广泛使用的一个助记口诀是LIATE，它给出了选取u的优先顺序：对数函数(Logarithmic)、反三角函数(Inverse trig)、代数函数(Algebraic)、三角函数(Trigonometric)、指数函数(Exponential)。列表中越早出现的函数通常应被选作u，因为其导数往往能简化剩下的积分。例如，在 ∫ x·ln x dx 中，对数出现在代数之前，因此设 u = ln x，dv = x dx。那么 du = (1/x) dx，v = x²/2，得到 x·ln x dx = (x²/2)ln x − ∫ (x²/2)(1/x) dx = (x²/2)ln x − ∫ x/2 dx = (x²/2)ln x − x²/4 + C。而在 ∫ x·eˣ dx 中，代数部分x在指数之前，所以选 u = x，dv = eˣ dx，最终得到 x·eˣ − ∫ eˣ dx = x·eˣ − eˣ + C。

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7. Repeated Integration by Parts | 多次分部积分

e integrals require applying the parts formula more than once. A classic example is ∫ x²·eˣ dx. First, set u = x², dv = eˣ dx ⇒ du = 2x dx, v = eˣ. This gives x²·eˣ − ∫ 2x·eˣ dx = x²·eˣ − 2∫ x·eˣ dx. The new integral ∫ x·eˣ dx itself needs parts: u = x, dv = eˣ dx ⇒ x·eˣ − ∫ eˣ dx = x·eˣ − eˣ. Putting it together, the final answer is x²·eˣ − 2(x·eˣ − eˣ) + C = eˣ o use algebraic manipulation after two rounds of parts to solve for the original integral.

有些积分需要多次使用分部公式。一个经典例子是 ∫ x²·eˣ dx。首先设 u = x²，dv = eˣ dx ⇒ du = 2x dx，v = eˣ。这得到 x²·eˣ − ∫ 2x·eˣ dx = x²·eˣ − 2∫ x·eˣ dx。新积分 ∫ x·eˣ dx 本身也需要分部：u = x，dv = eˣ dx ⇒ x·eˣ − ∫ eˣ dx = x·eˣ − eˣ。合并后，最终答案为 x²·eˣ − 2(x·eˣ − eˣ) + C = eˣ(x² − 2x + 2) + C。当同类积分再次出现时——比如指数与三角函数的乘积——你甚至可能需要在两轮分部之后使用代数技巧来解出原积分。

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8. Combining Substitution and Parts | 换元法与分部法的结合

problems ask you to use both techniques in succession. A typical question might be ∫ sin(√x) dx. Start with a substitution to remove the composite argument: let t = √x ∫ t sin t dt. Now this is a prime candidate for integration by parts: u = t, dv = sin t dt ⇒ us 2[ −t cos t − ∫ (−cos t) dt ] = 2[ −t cos t + sin t ] + C. Finally, substitute back t = √x to obtain 2[ −√x cos(√x) + sin(√x) ] + C. Always watch for opportunities to simplify the structure before applying parts.

较难的Edexcel题目会要求你依次使用两种技巧。一个典型问题是 ∫ sin(√x) dx。先用换元法消去复合参数：令 t = √x，则 x = t²，dx = 2t dt。积分变为 ∫ sin t · 2t dt = 2∫ t sin t dt。现在这成了分部积分的最佳候选：u = t，dv = sin t dt ⇒ du = dt，v = −cos t。于是 2[ −t cos t − ∫ (−cos t) dt ] = 2[ −t cos t + sin t ] + C。最后代回 t = √x 得到 2[ −√x cos(√x) + sin(√x) ] + C。在使用分部法之前，始终要留意是否存在简化结构的机会。

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9. Special Cases – Integrating the Natural Logarithm | 特例：对数函数的积分

dv = 1 dx, so du = (1/x) dx and v = x. The formula gives x·ln x − ∫ x·(1/x) dx = x·ln x − ∫ 1 dx = x·ln x − x + C. This trick is frequently assessed in Edexcel papers and is a good example of how parts can 'create' a simpler integral from one that looks unfamiliar.

ln x 的积分并不出现在标准列表中，但可以通过巧妙地应用分部法求出。把 ∫ ln x dx 写成 ∫ 1·ln x dx。现在设 u = ln x，dv = 1 dx，于是 du = (1/x) dx，v = x。公式给出 x·ln x − ∫ x·(1/x) dx = x·ln x − ∫ 1 dx = x·ln x − x + C。这个技巧在Edexcel试卷中经常出现，很好地展示了分部法如何从一个看似陌生的积分中“创造”出更简单的积分。

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10. Common Pitfalls and Exam Tips | 常见陷阱与应试技巧

ions. When using integration by parts, always check that the new integral ∫ v du is actually simpler than the original. If it looks worse, you probably chose u and dv the wrong way round. Also be careful with algebraic signs: a missing minus sign in trigonometric integrals is one of the most frequent errors. When a question says 'hence evaluate' or 'use the substitution', follow the instruction exactly – sets like pdfjoiner_(4)-093 under timed conditions builds the speed and accuracy you need.

许多学生因遗忘不定积分中的积分常数+C或在定积分换元时误用积分限而丢分。使用分部积分法时，务必检查新积分 ∫ v du 是否真的比原积分更简单。如果它看起来更复杂，你很可能选反了u和dv。同时要注意代数符号：三角积分中漏掉负号是最常见的错误之一。当题目中出现“由此计算”或“使用换元法”时，要严格遵循指令——考官希望看到解题方法，而不仅仅是光秃秃的答案。在定时条件下练习像pdfjoiner_(4)-093这样的题目集，能培养你所需的速度和准确性。

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11. Worked Synoptic Example | 综合例题详解

Find ∫ from 0 to π/2 of eˣ onential and trigonometric parts, we expect to use integration by parts twice. Set I = ∫ eˣ sin x dx. First time: let u = sin x, dv = eˣ dx ⇒ du = cos x dx, v = eˣ. Then I = eˣ sin x − ∫ eˣ cos x dx. For the new integral J = ∫ eˣ cos x dx, apply parts again: u = cos x, dv = eˣ dx ⇒ du = −sin x dx, v = eˣ. So J = eˣ cos x − ∫ eˣ(−sin x) dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I. Substituting back: I = eˣ sin x − (eˣ cos x + I) = eˣ sin x − eˣ cos x − I. Adding I to both sides gives 2I = eˣ(sin x − cos x), so I = (1/2)eˣ(sin x − cos x) + C. For the definite integral from 0 to π/2: evaluate [(1/2)eˣ(sin x − cos x)] at π/2 and 0. At x = π/2: (1/2)e^π/2(1 − 0) = (1/2)e^π/2. At x = 0: (1/2)(1)(0 − 1) = −1/2. Subtracting gives (1/2)e^π/2 − (−1/2) = (1/2)(e^π/2 + 1). This example illustrates the power of algebraic manipulation when an integral repeats.

求定积分 ∫ 从0到 π/2 的 eˣ sin x dx。被积函数同时包含指数与三角函数，我们预计需要使用两次分部积分。设 I = ∫ eˣ sin x dx。第一次：令 u = sin x，dv = eˣ dx ⇒ du = cos x dx，v = eˣ。那么 I = eˣ sin x − ∫ eˣ cos x dx。对于新积分 J = ∫ eˣ cos x dx，再次分部：u = cos x，dv = eˣ dx ⇒ du = −sin x dx，v = eˣ。于是 J = eˣ cos x − ∫ eˣ(−sin x) dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I。代回：I = eˣ sin x − (eˣ cos x + I) = eˣ sin x − eˣ cos x − I。两边加 I 得 2I = eˣ(sin x − cos x)，故 I = (1/2)eˣ(sin x − cos x) + C。对于从0到π/2的定积分：计算 [(1/2)eˣ(sin x − cos x)] 在π/2和0的值。x = π/2时：(1/2)e^π/2(1 − 0) = (1/2)e^π/2。x = 0时：(1/2)(1)(0 − 1) = −1/2。相减得 (1/2)e^π/2 − (−1/2) = (1/2)(e^π/2 + 1)。这个例子说明了当积分重现时代数操作的力量。

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12. Final Checks Before the Exam | 考前最后检查

ion, always show every step of your substitution or parts working. For substitution, write 'Let u = …' and 'du = …' before substituting. Keep an eye on the clock, but never rush the algebraic simplification – l to verify it returns the original integrand. This discipline not only catches slips but also deepens your understanding of the connection between differentiation and integration. With consistent practice, the techniques outlined here will become second nature, allowing you to approach questions like those in pdfjoiner_(4)-093 with confidence.

在A-Level考试中，始终要展示换元或分部积分的每一个步骤。对换元法，要清晰地写出“令 u = …”和“du = …”，并在定积分中变换积分限。对分部法，在代入前写下 u、dv、du、v 和完整公式。留意时间，但绝不要仓促进行代数化简——四分之一的分值可能因粗心的因式分解或符号错误而丢失。最后，如果你提前完成，可以对你的不定积分求导，验证它是否回到了原来的被积函数。这个习惯不仅能发现疏忽，还能加深你对微分与积分关系的理解。通过持续练习，本文所述的技巧将变成你的第二本能，让你能自信地应对pdfjoiner_(4)-093这类题目。

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