Integration by Substitution | 换元积分法

📚 Integration by Substitution | 换元积分法

Integration by substitution, often called u-substitution, is one of the most fundamental techniques for finding antiderivatives. It is the reverse process of the chain rule for differentiation. Whenever we see a composite function multiplied by the derivative of the inner function, substitution lets us rewrite the integral into a simpler form that we can evaluate directly. In IB Mathematics, mastering substitution enables you to tackle a wide range of integrals that appear in both Analysis & Approaches and Applications & Interpretation courses.

换元积分法，通常称为u代换，是寻找原函数最基本的方法之一，本质上是链式法则求导的逆运算。每当被积函数中出现复合函数乘以内层函数的导数时，通过代换可以将积分改写成可以直接计算的形式。掌握换元法能帮助你在IB数学中解决各种积分问题，无论是在分析与方法还是应用与解释课程中都经常出现。

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1. Reverse of the Chain Rule | 链式法则的逆运算

When we differentiate a composite function F(g(x)), the chain rule gives F'(g(x))·g'(x). Therefore, if we need to integrate an expression of the form f(g(x))·g'(x), we can recognise that it came from differentiating F(g(x)). Substitution makes this recognition systematic: we set u = g(x) so that du = g'(x) dx, and the integral ∫ f(g(x)) g'(x) dx becomes ∫ f(u) du, which is often far easier to handle.

当我们对复合函数F(g(x))求导时，链式法则给出F'(g(x))·g'(x)。因此，如果需要积分形如f(g(x))·g'(x)的表达式，我们可以识别出它来自对F(g(x))求导。代换法将这一识别过程系统化：令u = g(x)，则du = g'(x) dx，积分∫ f(g(x)) g'(x) dx就变成了∫ f(u) du，处理起来通常容易得多。

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2. The Substitution Procedure for Indefinite Integrals | 不定积分的代换步骤

1. Choose a new variable u = g(x) that simplifies the integrand. Look for an inner function whose derivative also appears in the integral.

1. 选取一个新变量 u = g(x) 以简化被积函数。寻找一个内层函数，且其导数也在积分中出现。

2. Compute du by differentiating: du = g'(x) dx. Solve for dx if necessary: dx = du / g'(x).

2. 通过求导计算 du：du = g'(x) dx。如果需要，可以解出 dx：dx = du / g'(x)。

3. Substitute both u and du into the original integral, replacing all x‑terms. The resulting integral should involve only u.

3. 将 u 和 du 同时代入原积分，消去所有关于 x 的项。得到的积分应只含有 u。

4. Evaluate the simpler integral in terms of u.

4. 计算关于 u 的简单积分。

5. Substitute back u = g(x) to express the antiderivative in terms of the original variable x. Do not forget the constant of integration + C.

5. 将 u = g(x) 代回，用原变量 x 表示原函数。不要忘记加上积分常数 + C。

∫ f(g(x))·g'(x) dx = ∫ f(u) du, where u = g(x)

∫ f(g(x))·g'(x) dx = ∫ f(u) du，其中 u = g(x)

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3. First Examples | 第一个例子

Example 1: Evaluate ∫ 2x·cos(x²) dx. We recognise the inner function x², whose derivative 2x appears exactly. Let u = x² ⇒ du = 2x dx. The integral becomes ∫ cos u du = sin u + C. Substituting back gives sin(x²) + C.

例1：计算 ∫ 2x·cos(x²) dx。我们识别出内层函数 x²，其导数 2x 恰好出现。令 u = x² ⇒ du = 2x dx。积分变为 ∫ cos u du = sin u + C。代回得到 sin(x²) + C。

Example 2: Evaluate ∫ (3x² + 2) √(x³ + 2x) dx. Let u = x³ + 2x ⇒ du = (3x² + 2) dx. The whole integrand becomes √u du = u^(1/2) du. Integration gives (2/3) u^(3/2) + C, so the answer is (2/3)(x³ + 2x)^(3/2) + C.

例2：计算 ∫ (3x² + 2) √(x³ + 2x) dx。令 u = x³ + 2x ⇒ du = (3x² + 2) dx。整个被积函数变为 √u du = u^(1/2) du。积分得 (2/3) u^(3/2) + C，答案为 (2/3)(x³ + 2x)^(3/2) + C。

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4. Recognising Suitable Substitutions | 识别合适的代换

Look for a function u = g(x) such that g'(x) is essentially present as a factor. In many IB problems, the numerator of a rational function is close to the derivative of the denominator, suggesting u = denominator. For integrands containing a linear function raised to a power, u = linear function often works. With exponentials, u = exponent when its derivative also shows up.

寻找一个函数 u = g(x)，使得 g'(x) 基本上作为因子出现。在很多IB问题中，有理函数的分子与分母的导数接近，这提示可设 u = 分母。对于含有线性函数的幂次，u = 线性函数通常有效。对于指数函数，当指数的导数也出现时，可设 u = 指数。

Common patterns:

常见模式：

∫ (f'(x) / f(x)) dx → u = f(x) gives ∫ (1/u) du = ln|u| + C.

∫ (f'(x) / f(x)) dx → 令 u = f(x) 得到 ∫ (1/u) du = ln|u| + C。

∫ f(ax + b) dx → u = ax + b, du = a dx, so dx = du/a.

∫ f(ax + b) dx → u = ax + b, du = a dx, 所以 dx = du/a。

∫ sinⁿ x cos x dx → u = sin x, du = cos x dx.

∫ sinⁿ x cos x dx → u = sin x, du = cos x dx。

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5. Linear Substitutions | 线性代换

For integrals of the form ∫ e^(ax+b) dx or ∫ (ax + b)ⁿ dx, a linear substitution simplifies the work. Let u = ax + b ⇒ du = a dx ⇒ dx = du/a. For instance, ∫ e^(5x‑2) dx becomes (1/5) ∫ e^u du = (1/5) e^u + C = (1/5) e^(5x‑2) + C. This shows that substituting a linear function only requires dividing by the coefficient a.

对于形如 ∫ e^(ax+b) dx 或 ∫ (ax + b)ⁿ dx 的积分，线性代换能简化计算。令 u = ax + b ⇒ du = a dx ⇒ dx = du/a。例如，∫ e^(5x‑2) dx 变为 (1/5) ∫ e^u du = (1/5) e^u + C = (1/5) e^(5x‑2) + C。这显示代换线性函数只需要除以系数 a 即可。

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6. Substitution in Definite Integrals | 定积分中的代换

When evaluating a definite integral using substitution, you have two choices: (i) substitute back to x before applying the original limits, or (ii) change the limits to u‑values and never return to x. The second method is usually cleaner. If u = g(x), when x = a, u = g(a); when x = b, u = g(b). The definite integral becomes ∫_{u=g(a)}^{u=g(b)} f(u) du.

在用代换法计算定积分时，有两种选择：(i) 在应用原积分限之前先代回 x，或者 (ii) 将积分限改为 u 值，再也不用回到 x。第二种方法通常更整洁。若 u = g(x)，当 x = a 时 u = g(a)；当 x = b 时 u = g(b)。定积分变为 ∫_{u=g(a)}^{u=g(b)} f(u) du。

Example: Evaluate ∫₀¹ 2x·(x² + 1)³ dx. Let u = x² + 1. When x = 0, u = 1; when x = 1, u = 2. Also du = 2x dx. The integral becomes ∫₁² u³ du = [u⁴/4]₁² = (16/4) – (1/4) = 15/4.

例题：计算 ∫₀¹ 2x·(x² + 1)³ dx。令 u = x² + 1。当 x = 0 时 u = 1；当 x = 1 时 u = 2。同时 du = 2x dx。积分变为 ∫₁² u³ du = [u⁴/4]₁² = (16/4) – (1/4) = 15/4。

Always remember to write your limits in terms of u when you perform the substitution. Using x‑limits on a u‑integral is a very common mistake.

务必在代换时将积分限写成 u 的形式。在 u 的积分上使用 x 的积分限是非常常见的错误。

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7. Trigonometric Substitutions | 三角代换

For integrands containing √(a² – x²), √(a² + x²) or √(x² – a²), ordinary u‑substitution may not be enough. A trigonometric substitution can transform the radical into a trigonometric expression. The most common case in IB is √(a² – x²): let x = a sin θ, then dx = a cos θ dθ and √(a² – x²) becomes a cos θ. Afterwards, you integrate with respect to θ and finally convert back to x using a right‑angled triangle or inverse trig functions.

对于含有 √(a² – x²)、√(a² + x²) 或 √(x² – a²) 的被积函数，普通的 u 代换可能不够。三角代换可以将根号化为三角表达式。IB中最常见的是 √(a² – x²)：令 x = a sin θ，则 dx = a cos θ dθ，√(a² – x²) 变为 a cos θ。之后对 θ 积分，最后用直角三角形或反三角函数换回 x。

Example: ∫ dx / √(1 – x²) with x = sin θ, dx = cos θ dθ, gives ∫ (cos θ / cos θ) dθ = ∫ 1 dθ = θ + C = arcsin x + C. This result is a standard formula, but the substitution justifies it.

例题：∫ dx / √(1 – x²)，设 x = sin θ, dx = cos θ dθ，得到 ∫ (cos θ / cos θ) dθ = ∫ 1 dθ = θ + C = arcsin x + C。这个结果是一个标准公式，但代换为其提供了合理性。

Trigonometric substitutions are only required at the HL level in Analysis & Approaches, but they illustrate the power of a well‑chosen substitution.

三角代换仅在分析与方法课程的高水平中要求，但它说明了恰当选择代换的强大之处。

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8. Working with Exponential and Logarithmic Functions | 指数和对数函数的代换

Exponential integrals often respond well to u = exponent. For ∫ e^(2x+1) dx, let u = 2x+1 ⇒ du = 2 dx, so dx = du/2. The result is (1/2) e^(2x+1) + C. When a logarithm is involved, such as ∫ (ln x)/x dx, set u = ln x ⇒ du = (1/x) dx, giving ∫ u du = (u²)/2 + C = (ln x)²/2 + C.

指数函数的积分常对 u = 指数 反应良好。对于 ∫ e^(2x+1) dx，令 u = 2x+1 ⇒ du = 2 dx，所以 dx = du/2。结果为 (1/2) e^(2x+1) + C。当涉及对数时，例如 ∫ (ln x)/x dx，设 u = ln x ⇒ du = (1/x) dx，得到 ∫ u du = (u²)/2 + C = (ln x)²/2 + C。

Another useful case is ∫ a^(kx) dx. Rewrite a^(kx) = e^(kx ln a) and use the substitution u = (k ln a) x, or directly remember that the derivative of a^(kx) involves ln a, so division by ln a k is needed.

另一个有用的情况是 ∫ a^(kx) dx。将其改写为 e^(kx ln a) 并使用代换 u = (k ln a) x，或直接记住 a^(kx) 的导数含有 ln a，因此需要除以 ln a 与 k 的乘积。

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9. Common Pitfalls and How to Avoid Them | 常见陷阱及避免方法

Mistake 1: Forgetting to replace dx entirely. After choosing u, you must express dx in terms of du. Simply writing du where dx was is insufficient. Always compute du = g'(x) dx and explicitly solve for dx if needed.

错误1：忘记完全替换 dx。选择 u 之后，必须用 du 表示 dx。仅仅用 du 代替 dx 是不够的。始终要计算 du = g'(x) dx，并在需要时明确解出 dx。

Mistake 2: Not adjusting the limits in a definite integral. If you change variables, the limits must change to u‑values. Leaving x‑limits will produce a wrong answer.

错误2：没有调整定积分的积分限。如果变量改变了，积分限必须变成 u 的值。保留 x 的积分限会产生错误答案。

Mistake 3: Substituting back too early in a complicated integral. Work entirely in u until you have a simple antiderivative, then substitute back.

错误3：在复杂积分中过早地代回。完全在 u 的世界中计算，直到得到简单的原函数，然后再代回。

Mistake 4: Forgetting the constant of integration in indefinite integrals. Always write + C at the final step after substituting back to x.

错误4：在不定积分中忘记积分常数。在代回 x 之后的最后一步务必写上 + C。

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10. Practice Strategy and Summary | 练习策略与总结

To become fluent, practise identifying the inner function by asking: “What function is inside another?” Start with problems where the derivative is an exact factor, then move to cases where you need to adjust by a constant. Gradually introduce definite integrals and trigonometric forms. Remember that substitution is essentially pattern recognition; the more examples you work through, the more natural the choices become.

要熟练运用，需要不断练习识别内层函数，问自己：“哪个函数在里面？”从导数是精确因子的题目入手，再过渡到需要乘以或除以常数的情形。逐步引入定积分和三角形式。记住，代换本质上是一种模式识别；你练习的例题越多，选择代换就越自然。

The key formula to keep in mind is:

需要牢记的核心公式是：

∫ f(g(x)) g'(x) dx = ∫ f(u) du, u = g(x)

∫ f(g(x)) g'(x) dx = ∫ f(u) du, u = g(x)

Integration by substitution transforms a seemingly difficult integral into a basic one by changing the variable of integration. With careful practice, you will be able to handle all the integration challenges in IB Mathematics confidently.

换元积分法通过改变积分变量，将看似困难的积分转化为基本积分。经过仔细练习，你将能够自信地应对IB数学中所有的积分挑战。

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