Mastering Integration by Substitution (Exercise 16E) | 掌握换元积分法（习题 16E）

📚 Mastering Integration by Substitution (Exercise 16E) | 掌握换元积分法（习题 16E）

Integration by substitution is one of the most powerful techniques for evaluating indefinite and definite integrals. In Exercise 16E, we focus on honing the skill of recognising the inner function and correctly applying the chain rule in reverse. Understanding when and how to use substitution is essential for success in IB Higher Level Mathematics and beyond.

换元积分法是求解不定积分和定积分最有效的方法之一。在习题 16E 中，我们将重点训练识别内层函数并正确逆用链式法则的能力。理解何时以及如何使用换元，对于在 IB 高等数学及更深入的学习中取得成功至关重要。

---

1. The Core Idea of Substitution | 换元法的核心思想

erses the chain rule for differentiation. If we have an integral of the form ∫ f(g(x)) · g' es it to ∫ f(u) du. The key is to spot a function and its derivative within the integrand.

换元法逆转了微分的链式法则。若被积函数形如 ∫ f(g(x)) · g'(x) dx，令 u = g(x) 可将其简化为 ∫ f(u) du。关键在于识别出被积函数中存在一个函数及其导数。

---

2. Choosing the Substitution u | 选择换元变量 u

The most common choice is to let u be the 'inner' x², so we set u = x². Then du = 2x dx, and the integral becomes ∫ cos(u) du = sin(u) + C.

最常见的选择是令 u 为复合函数的“内层”函数。例如，在 ∫ 2x·cos(x²) dx 中，我们注意到 2x 是 x² 的导数，因此设 u = x²。于是 du = 2x dx，积分化为 ∫ cos(u) du = sin(u) + C。

---

3. Adjusting for Constants | 调整常数因子

Sometimes the derivative is missing a constant multiplier. In ∫ x·e^3x² nce we only have x dx in the integrand, we solve for x dx = du/6. The integral becomes ∫ e^u (du/6) = (1/6)e^u + C.

有时导数缺了一个常数倍数。在 ∫ x·e^3x² dx 中，我们可设 u = 3x²，得 du = 6x dx。因为被积函数中只有 x dx，我们解出 x dx = du/6。积分化为 ∫ e^u (du/6) = (1/6)e^u + C。

---

4. Substitution with Definite Integrals | 定积分中的换元

t u = x²+1. When x=0, u=1; when x=1, u=2. The integral becomes ∫₁² u³ du = [u⁴/4]₁² = 16/4 – 1/4 = 15/4.

当积分带有限值时，我们可以将上下限转换为 u 值，或者在求值前换回 x。对于 ∫₀¹ 2x·(x²+1)³ dx，设 u = x²+1。当 x=0 时，u=1；当 x=1 时，u=2。积分化为 ∫₁² u³ du = [u⁴/4]₁² = 16/4 – 1/4 = 15/4。

---

5. Handling Trigonometric Integrals | 处理三角函数的积分

u³ du = u⁴/4 + C = (sin⁴x)/4 + C. Always look for a trig function and its derivative pair.

换元法对三角积分非常有效。在 ∫ sin³x cos x dx 中，令 u = sin x，则 du = cos x dx。积分化为 ∫ u³ du = u⁴/4 + C = (sin⁴x)/4 + C。要时刻留意是否存在三角函数及其导数对。

---

广告 / Sponsored

6. The 'ln' Form from Substitution | 由换元得到的 'ln' 形式

If an integrand has f'(x)/f(x), substitution u = f(x) gives ∫ du/u = ln|u| + C. For example, ∫ (2x+1)/(x²+x+5) dx: let u = x²+x+5, du = (2x+1) dx. The result is ln|x²+x+5| + C.

若被积函数为 f'(x)/f(x)，换元 u = f(x) 可得 ∫ du/u = ln|u| + C。例如，∫ (2x+1)/(x²+x+5) dx：令 u = x²+x+5，du = (2x+1) dx。结果为 ln|x²+x+5| + C。

---

7. Exponential Functions and Substitution | 指数函数与换元

Integrals involving e^ax+b titution, but careful handling is needed when the exponent is a more complex function. ∫ e^{sin x} cos x dx: set u = sin x, du = cos x dx, yielding ∫ e^u du = e^u + C = e^{sin x} + C.

涉及 e^ax+b 的积分常通过换元求解，但当指数更复杂时需谨慎处理。∫ e^{sin x} cos x dx：设 u = sin x，du = cos x dx，得到 ∫ e^u du = e^u + C = e^{sin x} + C。

---

8. Avoiding Common Pitfalls | 避免常见错误

One error is forgetting to replace dx with du/dx correctly. If u = g(x), then du = g' grand is expressed in terms of u before integrating. Also, never mix variables; if a stray 'x' remains after substitution, you may need a different approach.

一个常见错误是忘记将 dx 正确地替换为 du/dx。若 u = g(x)，则 du = g'(x) dx。务必确保整个被积函数在积分前都已用 u 表示。此外，切勿混合变量；如果换元后还残留一个多余的 'x'，可能需要另寻他法。

---

9. Advanced Substitution: Completing the Square | 高级换元：配方法

When the integrand contains √ax²+bx+c or a quadratic denominator that doesn' followed by a trigonometric or hyperbolic substitution can be the key. For instance, ∫ 1/√4x²+4x+5 dx: write 4x²+4x+5 = 4(x+½)² + 4, then set u = x+½.

当被积函数含有 √ax²+bx+c 或不易分解的二次分母时，先配方再使用三角换元或双曲换元往往是关键。例如，∫ 1/√4x²+4x+5 dx：将 4x²+4x+5 写成 4(x+½)² + 4，然后设 u = x+½。

---

10. Practice Strategies for Exercise 16E | 习题 16E 的练习策略

Begin by scanning each integral for a function and its derivative. Work systematically: identify the candidate u, compute du, adjust for constants, and rewrite the integral. After integrating, always differentiate your answer mentally to verify correctness. This builds fluency and reduces errors in exams.

先从扫描每个积分中的函数及其导数对开始。有条不紊地操作：确定候选 u，计算 du，调整常数，重写积分。积分完毕后，务必在心中对答案求导以验证正确性。这能提升熟练度并减少考试中的失误。

---

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）