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。

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

lacing 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)

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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。

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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。

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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。

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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 ∫ uⁿ du. For n ≠ -1, the power rule gives 1/2 · uⁿ⁺¹/(n+1) + C = uⁿ⁺¹/(2(n+1)) + C. For n = -1, the integral is 1/2 ∫ u^-1 du = 1/2 ln|u| luding 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 ∫ uⁿ du。当 n ≠ -1 时，幂法则给出 1/2 · uⁿ⁺¹/(n+1) + C = uⁿ⁺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。

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

hen 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。

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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 ∫ u⁵ du = 1/6 · u⁶ / 6 + C = u⁶ / 36 + C = (3x² + 2)⁶ / 36

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