What is Integration by Parts? 什么是分部积分法

What is Integration by Parts? 什么是分部积分法？

Integration by Parts is one of the most powerful techniques in A-Level Mathematics. It is essentially the "reverse" of the Product Rule for differentiation. If the product rule tells us how to differentiate the product of two functions, integration by parts tells us how to integrate that product.

分部积分法是A-Level数学中最强大的积分技巧之一。它本质上是微分乘积法则的"逆运算"。如果说乘积法则告诉我们如何对两个函数的乘积进行微分，那么分部积分法则告诉我们如何对这个乘积进行积分。

In your A-Level exam, you'll encounter integrals that cannot be solved by basic methods like substitution or inspection. When you see a product of two distinct types of functions — for example, x·eˣ, x·ln x, or x·sin x — you should immediately think: Integration by Parts.

在你的A-Level考试中，你会遇到无法用代入法或观察法等基本方法求解的积分。当你看到两种不同类型函数的乘积时——例如x·eˣ、x·ln x或x·sin x——你应该立刻想到：分部积分法。

The Formula 核心公式

The Integration by Parts formula is derived directly from the product rule. Let's review:

分部积分公式直接从乘积法则推导而来。回顾一下：

Product Rule 乘积法则：
d/dx [u(x)·v(x)] = u'(x)·v(x) + u(x)·v'(x)

Integrating both sides and rearranging gives us the famous formula 两边积分并整理得到著名的公式：

∫ u dv = uv − ∫ v du

In the A-Level context, we write this as 在A-Level的语境中，我们将其写为：

∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx

Where 其中：

• u — the function we choose to differentiate (我们选择微分的函数)
• dv/dx — the function we choose to integrate (我们选择积分的函数)
• du/dx — the derivative of u (u的导数)
• v — the integral of dv/dx (dv/dx的积分)

The LIATE Rule 如何选择u？LIATE法则

The most critical decision in Integration by Parts is choosing which function to set as u. A poor choice can make the integral more complicated rather than simpler. The LIATE rule is your best friend here:

分部积分法中最关键的决策是选择哪个函数作为u。错误的选择会让积分变得更复杂而不是更简单。LIATE法则会给你很好的指导：

L → I → A → T → E

Priority (High → Low) 优先级（高→低）：

L — Logarithmic 对数函数 (ln x, logₐx)

I — Inverse Trig 反三角函数 (arcsin x, arctan x)

A — Algebraic 代数函数 (xⁿ, polynomials 多项式)

T — Trigonometric 三角函数 (sin x, cos x, tan x)

E — Exponential 指数函数 (eˣ, aˣ)

Key principle 核心原则：Choose u from higher priority in LIATE. Higher-priority functions are harder to integrate — so we differentiate them instead. 选择LIATE中优先级较高的函数作为u。高优先级的函数更难积分——所以我们选择对它们进行微分而不是积分。

For A-Level, the most common cases are 对于A-Level考试，最常见的情况是：

• Polynomial × Exponential 多项式×指数：u = polynomial (A beats E)
• Polynomial × Trig 多项式×三角：u = polynomial (A beats T)
• Logarithmic × Anything 对数×任意：u = ln x (L beats everything)

Worked Example 1: x·eˣ 经典示例一

Let's integrate ∫ x·eˣ dx

这是A-Level中最经典的分布积分例题：

Step 1: Identify u and dv/dx 确定u和dv/dx
By LIATE: Algebraic (x) beats Exponential (eˣ) → u = x, dv/dx = eˣ

Step 2: Find du/dx and v 求导数和积分
du/dx = 1
v = ∫ eˣ dx = eˣ

Step 3: Apply the formula 代入公式
∫ x·eˣ dx = x·eˣ − ∫ eˣ·1 dx
= x·eˣ − ∫ eˣ dx
= x·eˣ − eˣ + C
= eˣ(x − 1) + C

Notice how the second integral became simpler — from ∫ x·eˣ dx we now only need ∫ eˣ dx. This is the hallmark of a good u choice. 注意第二个积分变得更简单了——从∫ x·eˣ dx我们只需要求∫ eˣ dx了。这就是正确选择u的标志。

Worked Example 2: x·sin x 经典示例二

Find ∫ x·sin x dx

Step 1: LIATE says u = x (Algebraic), dv/dx = sin x (Trigonometric)

Step 2:
du/dx = 1
v = ∫ sin x dx = −cos x

Step 3:
∫ x·sin x dx = x·(−cos x) − ∫ (−cos x)·1 dx
= −x·cos x + ∫ cos x dx
= −x·cos x + sin x + C
= sin x − x·cos x + C

Verification 验证： differentiate sin x − x·cos x:
cos x − [cos x + x·(−sin x)] = cos x − cos x + x·sin x = x·sin x ✓

Worked Example 3: ln x 对数函数积分（重要套路）

Integrate ∫ ln x dx

This is a special case — there's no visible product, but we can create one by writing ln x = 1·ln x. 这是一个特殊情况——没有明显的乘积，但我们可以写成ln x = 1·ln x。

Set 设：u = ln x, dv/dx = 1

du/dx = 1/x
v = ∫ 1 dx = x

∫ ln x dx = x·ln x − ∫ x·(1/x) dx
= x·ln x − ∫ 1 dx
= x·ln x − x + C

This is a mandatory result to memorise for A-Level — it appears frequently! 这个结果必须牢记，A-Level考试中经常出现！

Worked Example 4: x²·eˣ (Repeated Integration by Parts) 分部积分的重复使用

Sometimes one round of integration by parts isn't enough. When the "remaining integral" still contains a product, apply the method again. 有时一轮分部积分还不够。当"剩余积分"仍然包含乘积时，再次应用该方法。

Find ∫ x²·eˣ dx:

Round 1: u = x², dv/dx = eˣ
du/dx = 2x, v = eˣ

∫ x²·eˣ dx = x²·eˣ − ∫ 2x·eˣ dx
= x²·eˣ − 2∫ x·eˣ dx

Round 2 (using the result from Example 1 for ∫ x·eˣ dx):
= x²·eˣ − 2[x·eˣ − eˣ] + C
= x²·eˣ − 2x·eˣ + 2eˣ + C
= eˣ(x² − 2x + 2) + C

The "Loop" Case 循环型分部积分（高阶考点）

For integrals like ∫ eˣ·sin x dx, integration by parts creates a loop — you end up with the original integral again. 对于像∫ eˣ·sin x dx这样的积分，分部积分会形成一个循环——最终你会再次看到原来的积分。

Find ∫ eˣ·sin x dx:

Let u = sin x, dv/dx = eˣ
du/dx = cos x, v = eˣ

∫ eˣ·sin x dx = eˣ·sin x − ∫ eˣ·cos x dx

Apply again to ∫ eˣ·cos x dx: u = cos x, dv/dx = eˣ
du/dx = −sin x, v = eˣ

∫ eˣ·cos x dx = eˣ·cos x − ∫ eˣ·(−sin x) dx
= eˣ·cos x + ∫ eˣ·sin x dx

Now substitute back 代回：
∫ eˣ·sin x dx = eˣ·sin x − [eˣ·cos x + ∫ eˣ·sin x dx]
= eˣ·sin x − eˣ·cos x − ∫ eˣ·sin x dx

Bring the integral term to the left 把积分项移到左边：
2∫ eˣ·sin x dx = eˣ(sin x − cos x)
∫ eˣ·sin x dx = ½eˣ(sin x − cos x) + C

Exam Tips & Common Pitfalls 考试技巧与常见错误

✅ Do's 要做的：

• Always write out u, du/dx, dv/dx, v clearly — examiners award method marks 始终清晰地写出u、du/dx、dv/dx、v——阅卷老师会给方法分
• Check your answer by differentiating — it should give you the original integrand 通过微分验证答案——应该得到原来的被积函数
• Use LIATE to guide your choice of u 用LIATE法则指导u的选择
• For definite integrals, apply limits at the very end 对于定积分，在最后应用上下限

❌ Don'ts 不要做的：

• Don't forget the − sign in the formula 不要忘了公式中的负号
• Don't choose u based on "this part looks easier" — follow LIATE 不要基于"这部分看起来更简单"来选择u——遵循LIATE
• Don't forget +C for indefinite integrals 不定积分不要忘记+C
• Don't panic if the integral seems to get "more complicated" — trust the process 如果积分看起来变得更"复杂"，不要慌——信任这个过程

Quick Reference Card 速查卡

Save this for your revision 保存这个用于复习：

📐 Integration by Parts 分部积分法

Formula: ∫ u dv = uv − ∫ v du

LIATE Priority: Log → Inverse Trig → Algebraic → Trig → Exponential

Common results to memorise:
• ∫ x·eˣ dx = eˣ(x − 1) + C
• ∫ ln x dx = x(ln x − 1) + C
• ∫ x·sin x dx = sin x − x·cos x + C
• ∫ x·cos x dx = cos x + x·sin x + C
• ∫ eˣ·sin x dx = ½eˣ(sin x − cos x) + C

Practice Questions 练习题

Try these before your exam 考试前尝试这些：

1. ∫ x·ln x dx
2. ∫ x²·sin x dx
3. ∫ (ln x)² dx
4. ∫ x·e²ˣ dx (use substitution + IBP)
5. ∫ eˣ·cos 2x dx (the loop case)

Master these patterns and Integration by Parts will become second nature. Remember: the key is choosing the right u — and LIATE will never let you down.

掌握这些模式，分部积分法将变得轻而易举。记住：关键是选择正确的u——而LIATE法则永远不会让你失望。

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