Exercise 17D: Integration by Substitution | 练习17D：换元积分法

📚 Exercise 17D: Integration by Substitution | 练习17D：换元积分法

Mastering integration by substitution is a pivotal skill in IB Mathematics Analysis & Approaches. Exercise 17D typically focuses on transforming complex integrals into simpler forms using a change of variable. This article unpacks the technique step by step, with clear examples and common pitfalls, to help you tackle any substitution problem confidently.

掌握换元积分法是 IB 数学分析与方法中的关键技能。练习 17D 通常侧重于通过变量替换将复杂积分转化为简单形式。本文将逐步拆解这一技巧，配以清晰的示例和常见陷阱，帮助你自信地应对任何换元问题。

1. Introduction to Integration by Substitution | 换元积分法简介

Integration by substitution is the inverse process of the chain rule for differentiation. If an integrand can be recognised as the product of a composite function and the derivative of its inner function, a substitution u = g(x) will simplify the integral dramatically. The method transforms the variable of integration from x to u, making it possible to apply basic antiderivative formulas.

换元积分法是微分链式法则的逆过程。如果被积函数可以看作一个复合函数与其内部函数导数的乘积，那么使用 u = g(x) 进行替换将极大简化积分。该方法将积分变量从 x 转换为 u，使应用基本反导数公式成为可能。

2. The Reverse Chain Rule | 反链式法则

Recall the chain rule: d/dx [F(g(x))] = F'(g(x)) · g'(x). Integrating both sides gives ∫ F'(g(x)) · g'(x) dx = F(g(x)) + C. This is the foundation of substitution. When you spot a function and its derivative, set u = g(x) so that du = g'(x) dx. Then the integral becomes ∫ F'(u) du = F(u) + C.

回顾链式法则：d/dx [F(g(x))] = F'(g(x)) · g'(x)。对两边积分得到 ∫ F'(g(x)) · g'(x) dx = F(g(x)) + C。这就是换元法的基础。当你发现一个函数及其导数同时出现时，设 u = g(x)，于是 du = g'(x) dx。积分变为 ∫ F'(u) du = F(u) + C。

3. Choosing the Substitution u = g(x) | 选择替换 u = g(x)

The key to success is picking the right u. Usually, let u be the inner function of a composition, or an expression whose derivative also appears (perhaps up to a constant factor). For example, in ∫ 2x · sin(x²) dx, set u = x², du = 2x dx, and the integral becomes ∫ sin u du = -cos u + C = -cos(x²) + C. Constant multiples can be adjusted: ∫ x √(x²+1) dx, let u = x²+1, du = 2x dx, so x dx = ½ du, then the integral is ½ ∫ √u du.

成功的关键是选择正确的 u。通常设 u 为复合函数的内层函数，或者设为其导数也出现（可能相差一个常数倍）的表达式。例如，对于 ∫ 2x · sin(x²) dx，设 u = x²，du = 2x dx，积分变为 ∫ sin u du = -cos u + C = -cos(x²) + C。常数倍数可以调整：∫ x √(x²+1) dx，设 u = x²+1，du = 2x dx，所以 x dx = ½ du，积分变为 ½ ∫ √u du。

4. Substitution in Indefinite Integrals | 不定积分中的换元

When performing substitution on an indefinite integral, remember to express the final answer back in terms of the original variable x. After integrating with respect to u, replace u by g(x). Always include the constant of integration + C. Work systematically: identify u and du, rewrite the integral entirely in terms of u, integrate, then back-substitute.

对不定积分进行换元时，记住最终答案要换回原变量 x 表示。对 u 积分后，用 g(x) 替换 u。永远不要忘记加上积分常数 + C。系统地操作：确定 u 和 du，将积分完全用 u 重写，积分，然后回代。

5. Substitution in Definite Integrals | 定积分中的换元

For definite integrals, you have two options. Option 1: change the limits of integration to match the new variable u. If x ranges from a to b, then u ranges from g(a) to g(b). Then evaluate the u-integral directly without converting back to x. Option 2: keep the x-limits, find the antiderivative in terms of x as in the indefinite case, and evaluate using the original limits. Option 1 usually saves time and reduces errors.

对于定积分，你有两个选择。方法一：将积分限转换为新变量 u 的范围。如果 x 从 a 到 b，那么 u 从 g(a) 到 g(b)。然后直接计算 u 的定积分，无需换回 x。方法二：保留 x 的积分限，像不定积分那样求出原函数再代回原变量，用原限求值。方法一通常更省时且减少错误。

Example: ∫₀¹ 2x (x²+1)³ dx. Let u = x²+1, du = 2x dx. When x=0, u=1; when x=1, u=2. Integral = ∫₁² u³ du = [u⁴/4]₁² = (16/4)−(1/4) = 15/4.

示例：∫₀¹ 2x (x²+1)³ dx。设 u = x²+1，du = 2x dx。当 x=0 时 u=1；x=1 时 u=2。积分 = ∫₁² u³ du = [u⁴/4]₁² = (16/4)−(1/4) = 15/4。

6. Trigonometric Substitutions | 三角换元

When an integrand contains √(a²−x²), √(a²+x²), or √(x²−a²), trigonometric substitutions are powerful. For √(a²−x²), use x = a sin θ, dx = a cos θ dθ, and the root becomes a cos θ. For √(a²+x²), try x = a tan θ; then √(a²+x²) = a sec θ. For √(x²−a²), use x = a sec θ. Always remember to change the differential and the limits if definite, and back-substitute using right-triangle relationships if indefinite.

当被积函数包含 √(a²−x²)、√(a²+x²) 或 √(x²−a²) 时，三角换元非常有效。对于 √(a²−x²)，使用 x = a sin θ，dx = a cos θ dθ，根号部分变为 a cos θ。对于 √(a²+x²)，尝试 x = a tan θ，则 √(a²+x²) = a sec θ。对于 √(x²−a²)，使用 x = a sec θ。务必记得转换微分，若是定积分要改变上下限；若是不定积分，则利用直角三角形关系回代。

7. Rationalizing Substitutions | 有理化换元

Integrals involving roots of linear expressions, such as ∫ √(ax+b) dx or ∫ x/√(x+1) dx, can be handled by letting u be the entire radical expression or the inner linear function. For more complicated rational functions with different roots, a substitution like u = ∛(x) or u = x^(1/n) can eliminate fractional exponents. After substitution, the integrand often becomes a rational function of u, which can be integrated using partial fractions if necessary.

涉及线性表达式根式的积分，例如 ∫ √(ax+b) dx 或 ∫ x/√(x+1) dx，可以通过令 u 为整个根式或内部线性函数来处理。对于具有不同根式的更复杂有理函数，像 u = ∛(x) 或 u = x^(1/n) 这样的替换可以消除分数指数。替换后，被积函数通常变为 u 的有理函数，必要时可使用部分分式法积分。

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

One frequent mistake is forgetting to replace dx with du properly: you must express dx in terms of du, not simply substitute u. For example, if u = 2x+1, then du = 2 dx, so dx = du/2. Never write ∫ f(u) dx. Another error is not adjusting the limits for definite integrals or back-substituting incorrectly. Also, watch for hidden derivatives: sometimes the derivative of u differs from the given product by a constant factor, which you can pull outside the integral.

一个常见错误是忘记将 dx 正确替换为 du：必须用 du 表示 dx，而不是简单地替换 u。例如，若 u = 2x+1，则 du = 2 dx，所以 dx = du/2。决不能写成 ∫ f(u) dx。另一个错误是不调整定积分的上下限或回代错误。此外，注意隐藏的导数：有时 u 的导数与给定乘积相差一个常数倍，你可以将其提到积分外面。

9. Worked Example (IB Style) | 例题解析（IB风格）

Question: Find ∫ x √(2x+1) dx.

问题：求 ∫ x √(2x+1) dx。

Solution: Let u = 2x+1, then du = 2 dx, dx = ½ du. Also express x in terms of u: x = (u−1)/2. Substitute: ∫ x √(2x+1) dx = ∫ ((u−1)/2) · √u · (½ du) = ¼ ∫ (u−1) u^(1/2) du = ¼ ∫ (u^(3/2) − u^(1/2)) du. Integrate: ¼ [ (2/5)u^(5/2) − (2/3)u^(3/2) ] + C = (1/10)u^(5/2) − (1/6)u^(3/2) + C. Back-substitute u = 2x+1: = (1/10)(2x+1)^(5/2) − (1/6)(2x+1)^(3/2) + C. Factor if desired.

解答：令 u = 2x+1，则 du = 2 dx，dx = ½ du。同时将 x 用 u 表示：x = (u−1)/2。代入：∫ x √(2x+1) dx = ∫ ((u−1)/2) · √u · (½ du) = ¼ ∫ (u−1) u^(1/2) du = ¼ ∫ (u^(3/2) − u^(1/2)) du。积分得：¼ [ (2/5)u^(5/2) − (2/3)u^(3/2) ] + C = (1/10)u^(5/2) − (1/6)u^(3/2) + C。回代 u = 2x+1：= (1/10)(2x+1)^(5/2) − (1/6)(2x+1)^(3/2) + C。如有需要可提取公因子。

10. Practice Tips and Summary | 练习提示与总结

Begin by scanning the integrand for a composition of functions. Look for an inner function whose derivative appears alongside it. Practice with a variety of forms: exponential, logarithmic, trigonometric, and rational. Always check your answer by differentiating; the derivative should return the original integrand. For definite integrals, changing limits saves time, but double-check the new limits. With consistent practice, Exercise 17D will become a straightforward exercise in pattern recognition.

首先扫视被积函数，寻找复合函数。找出其内层函数，并检查其导数是否也出现在旁边。多加练习不同形式：指数、对数、三角和有理函数。始终通过求导来检验你的答案；导数应回到原被积函数。对于定积分，变换上下限可节省时间，但要仔细核对新限。通过持续练习，练习 17D 将变成一个简单的模式识别题。

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