Integration by Substitution | 掌握换元积分法

📚 Integration by Substitution | 掌握换元积分法

Integration by substitution is one of the most powerful reverse chain rule techniques in Edexcel A-Level Pure Mathematics. It transforms complex integrals by introducing a new variable to simplify the integrand, making it possible to tackle a wide range of functions that cannot be integrated directly. Mastery of this method is essential for the second year of the A-Level course and regularly appears on exam papers.

换元积分法是爱德思 A-Level 纯数学中最强大的逆链式法则技巧之一。它通过引入新变量来简化被积函数，将复杂积分转化为可处理的形式，从而能够计算大量无法直接积分的函数。掌握这种方法对 A-Level 第二年课程至关重要，也是考试中经常出现的题型。

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1. The Chain Rule in Reverse | 逆向链式法则

ution, recall the chain rule for differentiation: if y = f(u) and u = g(x), then dy/dx = f' ation by substitution reverses this process by recognising that an integrand of the form f'(g(x))·g'(x) can be integrated to f(g(x)) + C. This is why the method is often introduced as "integration by inspection" when the derivative of the inner function appears clearly.

要理解换元法，先回想微分中的链式法则：若 y = f(u) 且 u = g(x)，则 dy/dx = f'(u) × du/dx。换元积分法逆转了这一过程：当被积函数形如 f'(g(x))·g'(x) 时，可直接积出 f(g(x)) + C。因此，当内层函数的导数明显出现时，也常用“观察法”直接积分。

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2. The Formal Substitution Rule | 换元法则的正式形式

The general formula for integration by substitution is: ∫ f(g(x))·g' e all x-terms with u-terms. The differential dx is treated as a separate entity: du = (du/dx) dx, and the integral is entirely re-expressed in u before integrating.

换元积分法的一般公式为：∫ f(g(x))·g'(x) dx = ∫ f(u) du，其中 u = g(x)。实际操作中，我们选取被积函数中的某个表达式设为 u，对其求导得到 du/dx，然后将所有含 x 的项用 u 替换。微分 dx 被视为独立量，即 du = (du/dx) dx，积分完全转换为关于 u 的积分后再进行运算。

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3. Choosing the Right Substitution | 如何选择合适的换元

Deciding on an effective substitution is a skill developed through practice. Generally, look for a function whose derivative appears elsewhere in the integrand. Common choices include: the inner function of a composite, the denominator of a rational expression, the argument of a trigonometric function, or the expression under a root. Setting u equal to the "most complicated part" or the part inside brackets often works well.

选择合适的换元是一项通过练习培养的技能。通常，我们要寻找其导数在积分中其他部分出现的函数。常见选择包括：复合函数的内层、有理式的分母、三角函数的参数，或根号内的表达式。将 u 设为“最复杂的部分”或括号内的部分通常效果良好。

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4. Basic Polynomial Substitutions | 基本多项式换元

) u⁶ + C. Substituting back gives (1/6)(x² + 1)⁶ + C. Always verify by differentiating the result using the chain rule to check it yields the original integrand.

考虑积分 ∫ 2x(x² + 1)⁵ dx。设 u = x² + 1，则 du/dx = 2x，于是 du = 2x dx。积分变为 ∫ u⁵ du = (1/6) u⁶ + C。再将 u 代回得到 (1/6)(x² + 1)⁶ + C。务必用链式法则对结果求导，验证是否能得到原来的被积函数。

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5. Substitution with Definite Integrals | 定积分的换元法

he original limits. Changing limits is often more efficient: if u = g(x), then when x = a, u = g(a); when x = b, u = g(b). Integrate with respect to u and evaluate at the new limits—no need to substitute back.

用换元法计算定积分时，必须将积分限也转换到新变量，或者在积出原函数后代回 x 再使用原来的限。转换积分限通常更高效：如果 u = g(x)，那么当 x = a 时 u = g(a)，当 x = b 时 u = g(b)。直接对 u 积分并用新限求值，无需再代回 x。

Example: Evaluate ∫₀¹ 3x²√x³+1 Let u = x³+1, du/dx = 3x². When x = 0, u = 1; when x = 1, u = 2. The integral becomes ∫₁² √u du = [ (2/3) u^3/2 ]₁² = (2/3)(2√2 − 1). Correct handling of limits avoids algebraic errors.

例题：计算 ∫₀¹ 3x²√x³+1 dx。设 u = x³+1，du/dx = 3x²。当 x = 0 时 u = 1；x = 1 时 u = 2。积分变为 ∫₁² √u du = [ (2/3) u^3/2 ]₁² = (2/3)(2√2 − 1)。正确处理积分限可以避免代数错误。

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6. Trigonometric Substitutions | 三角换元

Trigonometric substitutions are vital for integrals involving √a²−x², √a²+x² or √x²−a². For √a²−x², use x = a sin θ, so dx = a cos θ dθ and the root simplifies to a cos θ. For √a²+x² xploit Pythagorean identities to eliminate the square root, allowing you to integrate using standard trigonometric integrals.

三角换元对含有 √a²−x²、√a²+x² 或 √x²−a² 的积分至关重要。对于 √a²−x²，令 x = a sin θ，则 dx = a cos θ dθ，根号简化为 a cos θ。对于 √a²+x²，可尝试 x = a tan θ。这些换元利用毕达哥拉斯恒等式消去根号，使积分可以用标准三角积分公式解决。

s result is a standard A-Level integral and appears frequently in partial fractions and differential equations.

例如，∫ 1/(1+x²) dx 可使用 x = tan θ，得出 dx = sec² θ dθ。积分变为 ∫ (sec² θ)/(1+tan² θ) dθ = ∫ 1 dθ = θ + C = arctan x + C。这是 A-Level 标准积分，经常出现在部分分式和微分方程中。

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7. Substitution by Recognising the Derivative | 识别导数型换元

the numerator is exactly the derivative of the denominator. For integrals of the form ∫ f'(x)/f(x) dx, the substitution u = f(x) yields ∫ (1/u) du = ln|u| + C. Hence, ∫ (2x)/(x²+1) dx = ln|x²+1| me and is heavily examined in Edexcel papers under the natural logarithm integration rule.

一个特殊情况是分子恰好为分母的导数。对于形如 ∫ f'(x)/f(x) dx 的积分，令 u = f(x) 就得到 ∫ (1/u) du = ln|u| + C。因此 ∫ (2x)/(x²+1) dx = ln|x²+1| + C。识别这一模式可以节省时间，也是爱德思考试中对数积分规则的重要考点。

Similarly, for integrands like f'(x)·e^f(x), substituting u = f(x) gives ∫ e^u du = e^f(x) + C. Example: ∫ 3x² e^x³ dx = e^x³ + C. These "derivative present" integrals are the cornerstone of rapid integration.

同样，对被积函数形如 f'(x)·e^f(x) 的情况，代换 u = f(x) 给出 ∫ e^u du = e^f(x) + C。例如 ∫ 3x² e^x³ dx = e^x³ + C。这类“导数出现”型积分是快速积分的基石。

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8. Integration of Rational Functions Using Substitution | 有理函数的换元积分

x+1, then du/dx = 2, so dx = du/2. The integral becomes ∫ (1/u³)·(1/2) du = (1/2) ∫ u⁻³ du = (1/2)(−1/2)u⁻² + C = −1/(4(2x+1)²) + C. Adjust the constants carefully.

分母为线性函数幂次的有理函数可以通过设 u 等于分母来积分。例如 ∫ 1/(2x+1)³ dx。令 u = 2x+1，则 du/dx = 2，dx = du/2。积分变为 ∫ (1/u³)·(1/2) du = (1/2) ∫ u⁻³ du = (1/2)(−1/2)u⁻² + C = −1/(4(2x+1)²) + C。要仔细处理常数。

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9. Handling Integrals Involving e^x and ln x | 处理含 eˣ 与 ln x 的积分

Substituting u = e^x or u = ln x often simplifies transcendental integrals. For ∫ e^x / (1+e^x) dx, let u = 1+e^x, du = e^x dx, giving ∫ 1/u du = ln|1+e^x| lting in ∫ u du = (1/2)(ln x)² + C. These are standard approaches for A-Level integration.

换元 u = e^x 或 u = ln x 常能简化超越函数积分。对于 ∫ e^x / (1+e^x) dx，令 u = 1+e^x，du = e^x dx，得到 ∫ 1/u du = ln|1+e^x| + C。对于 ∫ (ln x)/x dx，设 u = ln x，du = (1/x) dx，积分变为 ∫ u du = (1/2)(ln x)² + C。这是 A-Level 积分中的标准方法。

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10. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Major pitfalls include forgetting to change the dx appropriately, mishandling definite limits, or not substituting back to the original variable when giving the final answer for an indefinite integral. Never write an answer in terms of u for a question posed in x. Also, double-check algebraic simplifications—especially when dealing with coefficients that appear from du/dx.

主要陷阱包括忘记正确处理 dx 转换、定积分限处理不当，或不定积分最终答案未代回原变量。绝对不要对以 x 提问的题目给出以 u 表示的答案。此外，要仔细检查代数化简——尤其是处理来自 du/dx 的系数时。

Another error is choosing a substitution that does not simplify the integral. If after substitution the integral looks more complicated, try a different u or consider integration by parts. Practice with a variety of functions builds an instinct for effective substitutions.

另一个错误是选取的换元并没有简化积分。若换元后积分变得更复杂，应尝试换另一个 u 或考虑分部积分法。通过对各类函数的练习，可以培养选取有效换元的直觉。

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11. Worked Example with Detailed Steps | 步骤详细的例题演示

Let's integrate ∫ x√2x+1 ion. Set u = 2x+1. Then x = (u−1)/2 and du/dx = 2, so dx = du/2. Substitute: ∫ ((u−1)/2) · √u · (du/2) = (1/4) ∫ (u−1) u^1/2 du = (1/4) ∫ (u^3/2 − u^1/2) du. Integrate: (1/4)[ (2/5)u^5/2 − (2/3)u^3/2 ] + C = (1/10)(2x+1)^5/2 − (1/6)(2x+1)^3/2 + C.

我们来积分 ∫ x√2x+1 dx。设 u = 2x+1，则 x = (u−1)/2，且 du/dx = 2，dx = du/2。代入：∫ ((u−1)/2) · √u · (du/2) = (1/4) ∫ (u−1) u^1/2 du = (1/4) ∫ (u^3/2 − u^1/2) du。积分得出：(1/4)[ (2/5)u^5/2 − (2/3)u^3/2 ] + C = (1/10)(2x+1)^5/2 − (1/6)(2x+1)^3/2 + C。

This example shows how substitution can cope with a product of x and a root that is not simply the derivative of the inner function. The key is to express x in terms of u as well.

本例展示了换元法如何处理 x 与根式的乘积，即使被积函数并不单纯是内层函数的导数形式。关键是要将 x 也用 u 表示。

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12. Summary and Exam Tips | 总结与考试技巧

Integration by substitution is a core competency for Edexcel A-Level Maths. Remember: identify an inner function, find du/dx, replace dx, transform limits for definite integrals, and always present the final answer in terms of the original variable. In timed conditions, write the substitution and differential clearly to avoid losing method marks. Practise a wide range, from simple polynomials to complex trig and exponential forms, to ensure fluency.

换元积分法是爱德思 A-Level 数学的核心能力。记住：找出内层函数，求 du/dx，替换 dx，对定积分要转换积分限，并始终以原变量给出最终答案。在限时考试中，清晰写出换元与微分步骤，以避免丢失方法分。通过从简单多项式到复杂三角与指数形式的广泛练习，确保熟练掌握。

Quick Reference	Substitution Strategy
f'(x)/f(x)	u = f(x), result ln|f(x)|
f'(x)·ef(x)	u = f(x), result ef(x)
∫ k·f'(x)[f(x)]ⁿ dx	u = f(x), result k/(n+1) [f(x)]ⁿ⁺¹
√a²−x²	x = a sin θ
√a²+x²	x = a tan θ
Definite integral	Change limits to u-values

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