IB Mathematics: Integration by Substitution for x·f(ax²+b) | IB数学：ax²+b型复合函数乘x的换元积分法

📚 IB Mathematics: Integration by Substitution for x·f(ax²+b) | IB数学：ax²+b型复合函数乘x的换元积分法

In IB Mathematics, integration by substitution is one of the most useful techniques for finding antiderivatives. This article focuses on a very common pattern: the integrand is a product of x and a composite function of the form f(ax² + b). With the right substitution, these integrals can be evaluated in just a few steps.

在 IB 数学中，换元积分法是最有用的求原函数技巧之一。本文聚焦一个非常常见的题型：被积函数是 x 与 f(ax² + b) 型复合函数的乘积。使用正确的换元，这类积分只需几步就能求出。

1. The Core Substitution: u = ax² + b | 核心换元：u = ax² + b

The expression ax² + b appears as the inner function of a composite function, and it is multiplied by x. This is a classic pattern for integration by substitution. We set u = ax² + b. Then du = 2ax dx, which means x dx = du / (2a). Replacing x and dx in the original integral transforms it into 1/(2a) times the integral of f(u) with respect to u. This simplification is the reason the method is so powerful.

当被积函数是 x 乘以某个复合函数 f(ax² + b) 时，我们通常令 u = ax² + b。这是因为 du = 2ax dx，于是 x dx = du/(2a)。将原积分中的 x 和 dx 替换后，积分就化为 1/(2a) ∫ f(u) du。这种换元能够把复杂的复合函数积分转化为对 u 的简单积分。

u = ax² + b, du = 2ax dx, x dx = du / (2a)

2. Why the ‘x’ Factor is Essential | 为什么必须要有 x 因子

The derivative of ax² + b is 2ax, which contains both a constant factor 2a and the variable factor x. In the integral, the x next to f(ax² + b) supplies the missing variable part, leaving only a constant factor that can be moved outside the integral. If the x factor were absent, the differential du would not align with the integrand, and this simple substitution would not work. Recognizing this pattern is a key skill for IB exam questions.

ax² + b 的导数为 2ax，其中既含有常数因子 2a，也含有变量因子 x。积分中位于 f(ax² + b) 旁边的 x 恰好提供了缺失的变量部分，剩下的常数因子可以提到积分号外。如果缺少 x 因子，du 就无法与被积函数对齐，这种简单的换元也就无法使用。识别这种结构是 IB 考试中非常重要的能力。

3. Worked Example 1: ∫ x e^(x²) dx | 例1：∫ x e^(x²) dx

Let u = x². Then du = 2x dx, so x dx = du / 2. The integral becomes 1/2 ∫ e^u du. Since the integral of e^u is e^u, we obtain 1/2 e^u + C. Substituting u = x² back gives 1/2 e^(x²) + C. To verify, differentiate 1/2 e^(x²): d/dx [1/2 e^(x²)] = 1/2 e^(x²) · 2x = x e^(x²), which matches the integrand.

令 u = x²，则 du = 2x dx，所以 x dx = du/2。原积分变为 1/2 ∫ e^u du。因为 e^u 的积分是 e^u，所以我们得到 1/2 e^u + C。将 u = x² 代回，得 1/2 e^(x²) + C。验证：对 1/2 e^(x²) 求导，d/dx [1/2 e^(x²)] = 1/2 e^(x²) · 2x = x e^(x²)，与原被积函数一致。

4. Worked Example 2: ∫ x / √(x² + 4) dx | 例2：∫ x / √(x² + 4) dx

Choose u = x² + 4. Then du = 2x dx, and x dx = du / 2. The denominator becomes √u = u^(1/2). Thus the integral is 1/2 ∫ u^(-1/2) du. Using the power rule, ∫ u^(-1/2) du = 2 u^(1/2). Therefore the result is 1/2 · 2 u^(1/2) + C = u^(1/2) + C = √(x² + 4) + C. Notice that the +4 inside the square root does not complicate the integration; it is absorbed into the substitution.

取 u = x² + 4，则 du = 2x dx，得 x dx = du/2。分母变成 √u = u^(1/2)。因此积分为 1/2 ∫ u^(-1/2) du。根据幂法则，∫ u^(-1/2) du = 2 u^(1/2)。所以结果为 1/2 · 2 u^(1/2) + C = u^(1/2) + C = √(x² + 4) + C。注意根号内的 +4 并不会使积分变复杂，它会被换元过程完全吸收。

5. Worked Example 3: ∫ x (x² + 1)^n dx | 例3：∫ x (x² + 1)^n dx

Let u = x² + 1, du = 2x dx, so x dx = du / 2. The integral becomes 1/2 ∫ u^n du. For n ≠ -1, the power rule gives 1/2 · u^(n+1)/(n+1) + C = u^(n+1)/(2(n+1)) + C. For n = -1, the integral is 1/2 ∫ u^(-1) du = 1/2 ln|u| + C. This example shows the same substitution works for a whole family of powers, including negative powers. In IB questions, n may be a specific integer or a rational number.

令 u = x² + 1，du = 2x dx，所以 x dx = du/2。积分变为 1/2 ∫ u^n du。当 n ≠ -1 时，幂法则给出 1/2 · u^(n+1)/(n+1) + C = u^(n+1)/(2(n+1)) + C。当 n = -1 时，积分为 1/2 ∫ u^(-1) du = 1/2 ln|u| + C。这个例子说明同一种换元可以处理一族幂函数，包括负幂。在 IB 题目中，n 可能是具体整数或有理数。

6. Definite Integrals: Changing Limits | 定积分：换限

For a definite integral, after the substitution u = ax² + b, the original limits on x must be changed into limits on u. If the lower limit is x = L and the upper limit is x = U, then the new lower limit is u(L) and the new upper limit is u(U). Let us evaluate ∫₀¹ x / (x² + 1) dx. Set u = x² + 1, du = 2x dx. When x = 0, u = 1; when x = 1, u = 2. The integral becomes 1/2 ∫₁² (1/u) du = 1/2 [ln u]₁² = 1/2 (ln 2 – ln 1) = 1/2 ln 2. Notice that we never need to express the answer in terms of x for a definite integral.

对于定积分，在换元 u = ax² + b 之后，原来关于 x 的上下限必须换成关于 u 的上下限。若下限为 x = L，上限为 x = U，则新的下限为 u(L)，新的上限为 u(U)。我们计算 ∫₀¹ x/(x² + 1) dx。令 u = x² + 1，du = 2x dx。当 x = 0 时，u = 1；当 x = 1 时，u = 2。积分变为 1/2 ∫₁² (1/u) du = 1/2 [ln u]₁² = 1/2 (ln 2 – ln 1) = 1/2 ln 2。注意对于定积分，我们不需要把结果代回 x 的表达式。

7. More Complex Examples: Polynomial Factors | 更复杂的例子：多项式因子

The same idea extends to ∫ x (3x² + 2)^5 dx. Let u = 3x² + 2, so du = 6x dx. Then x dx = du / 6. The integral is 1/6 ∫ u^5 du = 1/6 · u^6 / 6 + C = u^6 / 36 + C = (3x² + 2)^6 / 36

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