Integration by Substitution & Parts | 换元积分法与分部积分法

📚 Integration by Substitution & Parts | 换元积分法与分部积分法

Mastering integration techniques is essential for A-Level Edexcel Mathematics, and two of the most powerful methods are integration by substitution and integration by parts. These techniques allow you to solve a wide variety of integrals that cannot be tackled directly using basic rules. This article provides a thorough revision of both methods, complete with step-by-step examples and common pitfalls to avoid.

掌握积分技巧是 A-Level Edexcel 数学的核心要求，而两种最有力的方法就是换元积分法和分部积分法。这些技巧能帮助你解决大量无法直接用基本积分公式求解的积分。本文将对这两种方法进行全面复习，提供分步演示的例题并指出需要避免的常见错误。

1. The Chain Rule in Reverse | 反向链式法则

of the chain rule for differentiation. When you spot an integral of the form ∫ f(g(x)) · g'(x) dx, you can simplify it by letting u = g(x). Then du = g' en much easier to evaluate. The key skill is identifying the inner function to substitute.

换元积分法本质上是微分中链式法则的逆运算。当你发现积分形如 ∫ f(g(x)) · g'(x) dx 时，可以通过令 u = g(x) 来简化。于是 du = g'(x) dx，积分就变成了 ∫ f(u) du，这样通常更容易计算。关键在于识别出需要替换的内层函数。

2. Choosing the Right Substitution | 选择合适的换元

vided, look for a function whose derivative also appears as a factor. For example, in ∫ 2x·√x²+1 natural choice. Always remember to change the limits if dealing with a definite integral.

对许多积分来说，考题会直接给出换元，尤其是涉及三角函数或指数表达式时。如果没有给出，就要寻找导数也作为因式出现的函数。例如在 ∫ 2x·√x²+1 dx 中，x²+1 的导数是 2x，因此很自然地选择 u = x²+1。如果是定积分，别忘了还要改变积分的上下限。

3. Step-by-Step Substitution Example | 换元积分分步例题

², then du = 2x dx, so x dx = ½ du. The integral becomes ∫ sin(u) · ½ du = ½ ∫ sin(u) du = –½ cos(u) + C. Substituting back gives –½ cos(x²) + C. Always check your answer by differentiating: the derivative of –½ cos(x²) is –½ · (–sin(x²)) · 2x = x sin(x²), confirming the result.

考虑 ∫ x·sin(x²) dx。令 u = x²，则 du = 2x dx，所以 x dx = ½ du。积分变为 ∫ sin(u) · ½ du = ½ ∫ sin(u) du = –½ cos(u) + C。回代后得到 –½ cos(x²) + C。一定要通过求导来检验你的答案：–½ cos(x²) 的导数是 –½ · (–sin(x²)) · 2x = x sin(x²)，从而验证结果正确。

4. Definite Integrals and Changing Limits | 定积分与变换上下限

erms of x and revert to the original variable before applying the original limits. The first method is usually cleaner. For instance, ∫ from 0 to √π/2 ∫ from 0 to π/2 of ½ sin(u) du with new limits found by substituting x-values into u = x².

在用换元法计算定积分时，有两种选择：在积分之前先将上下限换成 u 的值，或者先以 x 为变量积分，然后换回原变量再代入原始上下限。第一种方法通常更简洁。例如 ∫ 从 0 到 √π/2 x·sin(x²) dx 变为 ∫ 从 0 到 π/2 ½ sin(u) du，新上下限是通过把 x 的值代入 u = x² 得到的。

5. Integration by Parts – The Product Rule Reversed | 分部积分法——反向乘法法则

differentiation and is used to integrate products of functions. The formula is ∫ u dv = uv – grated easily. A common mnemonic is LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) to prioritise the choice of u.

分部积分法源自微分的乘法法则，用于计算两个函数乘积的积分。公式为 ∫ u dv = uv – ∫ v du。这里的 u 和 dv 是从原被积函数中选取的：选择 u 时要看它求导后能否简化，而选择 dv 时要看它是否容易积分。常用的助记口诀是 LIATE（对数函数、反三角函数、代数函数、三角函数、指数函数），用于确定选取 u 的优先级。

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6. Applying Integration by Parts Once | 应用一次分部积分法

To evaluate ∫ x·eˣ dx, set u = x (algebraic) and dv = eˣ dx (exponential). Then du = dx, v = eˣ. The formula gives x·eˣ – ∫ eˣ dx = x·eˣ – eˣ + C = eˣ(x – the algebraic term disappears after differentiation, leaving a basic exponential integral.

要计算 ∫ x·eˣ dx，设 u = x（代数函数），dv = eˣ dx（指数函数）。则 du = dx，v = eˣ。代入公式得 x·eˣ – ∫ eˣ dx = x·eˣ – eˣ + C = eˣ(x – 1) + C。这个例子说明了核心思想：乘积变简单了，因为代数项求导后消失，留下一个基本的指数积分。

7. Repeated Integration by Parts | 重复分部积分法

once. For ∫ x²·sin(x) dx, first let u = x², dv = sin(x) dx. After one application, you get – l still has a product, so use parts again with u = 2x, dv = cos(x) dx. The final result is –x²·cos(x) + 2x·sin(x) + 2cos(x) + C. Keep track of signs carefully.

有时你需要多次应用分部积分法。对于 ∫ x²·sin(x) dx，先令 u = x²，dv = sin(x) dx。应用一次后得到 –x²·cos(x) + ∫ 2x·cos(x) dx。这个新积分仍然是乘积，所以要再次使用分部积分法，设 u = 2x，dv = cos(x) dx。最终结果是 –x²·cos(x) + 2x·sin(x) + 2cos(x) + C。要小心处理符号。

8. The Special Case of ∫ ln(x) dx | 特例：∫ ln(x) dx

∫ 1·ln(x) dx and let u = ln(x), dv = 1 dx. Then du = (1/x) dx, v = x. This yields x·ln(x) – ∫ x·(1/x) dx = x·ln(x) – x + C. This trick also works for inverse trig functions like arcsin(x) or arctan(x).

即使被积函数只是单一的对数函数，也可以使用分部积分法。把 ∫ ln(x) dx 写成 ∫ 1·ln(x) dx，设 u = ln(x)，dv = 1 dx。则 du = (1/x) dx，v = x。于是得到 x·ln(x) – ∫ x·(1/x) dx = x·ln(x) – x + C。这个技巧对反正弦 arcsin(x) 或反正切 arctan(x) 等反三角函数也同样适用。

9. Combining Substitution and Parts | 换元法与分部积分法的结合

Some integrals require both techniques. For example, ∫ x³·e^x² is best approached by first using substitution: let w = x², dw = 2x dx, giving (1/2) ∫ w·eʷ aking a sensible initial substitution to reduce the integral to a form suitable for parts.

有些积分需要同时使用两种技巧。例如 ∫ x³·e^x² dx 最好先用换元法：令 w = x²，dw = 2x dx，得到 (1/2) ∫ w·eʷ dw。然后再对以 w 为变量的积分使用分部积分法。解决复杂的 A-Level 问题时常需要先做一个合理的换元，将积分转化为适合分部积分的形式。

10. Common Mistakes and Exam Tips | 常见错误与考试建议

For substitution, never forget to replace dx with du properly – a missing factor is a frequent error. For parts, ensure you choose u and dv wisely; picking u incorrectly can make the integral more complicated. In definite integrals, change the limits immediately after substitution to avoid confusion. Always check your final answer by differentiating when time permits.

在用换元法时，千万不要忘记将 dx 正确地替换成 du——漏掉一个系数是常见的错误。在用分部积分法时，要确保合理选择 u 和 dv；选错 u 可能会让积分变得更复杂。在定积分中，换元之后应立即更换上下限以避免混淆。如果时间允许，一定要通过求导来检查最终答案。

These two techniques form the backbone of A-Level integration and appear frequently in both pure and applied mathematics contexts. Practice recognising the structure of integrals and applying the appropriate method fluently, and you will build the confidence needed for exam success.

这两大技巧构成了 A-Level 积分学的支柱，并频繁出现在纯数学和应用数学的背景中。通过练习识别积分的结构并流畅地选择适当方法，你将建立起在考试中取得成功所需的信心。

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