📚 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。
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' ng 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。
2. The Substitution Procedure for Indefinite Integrals | 不定积分的代换步骤
implifies 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。
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)
3. First Examples | 第一个例子
²) dx. We recognise the inner function x², whose derivative 2x appears exactly. Let u = x² ⇒ = 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 = u1/2 du. Integration gives (2/3) u3/2 + 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 = u1/2 du。积分得 (2/3) u3/2 + C，答案为 (2/3)(x³ + 2x)3/2 + C。
4. Recognising Suitable Substitutions | 识别合适的代换
Look for a function u = g(x) such that g' r 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。
5. Linear Substitutions | 线性代换
For integrals of the form ∫ eax+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, ∫ e5x‑2 dx becomes (1/5) ∫ eu du = (1/5) eu + C = (1/5) e5x‑2 + C. This shows that substituting a linear function only requires dividing by the coefficient a.
对于形如 ∫ eax+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，∫ e5x‑2 dx 变为 (1/5) ∫ eu du = (1/5) eu + C = (1/5) e5x‑2 + C。这显示代换线性函数只需要除以系数 a 即可。
6. Substitution in Definite Integrals | 定积分中的代换
) substitute back to x before applying the original limits, or (ii) change the limits to u‑ . 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。
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。
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 + 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。
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 ∫ e2x+1 dx, let u = 2x+1 ⇒ du = 2 dx, so dx = du/2. The result is (1/2) e2x+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 ∫ e2x+1 dx，令 u = 2x+1 ⇒ du = 2 dx，所以 dx = du/2。结果为 (1/2) e2x+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 ∫ akx dx. Rewrite akx = ekx ln a and use the substitution u = (k ln a) x, or directly remember that the derivative of akx involves ln a, so division by ln a k is needed.
另一个有用的情况是 ∫ akx dx。将其改写为 ekx ln a 并使用代换 u = (k ln a) x，或直接记住 akx 的导数含有 ln a，因此需要除以 ln a 与 k 的乘积。
9. Common Pitfalls and How to Avoid Them | 常见陷阱及避免方法
s 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。
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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