Integrating Using Integration by Parts · 分部积分法积分
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| Integration by parts/ˌɪntɪˈɡreɪʃn baɪ pɑːts/ | 分部积分法 | fēn bù jī fēn fǎ |
The product rule, run backward
- The chain rule reversed gave u-substitution. What reverses the product rule? Integration by parts 分部积分法.
- It handles integrals of a product that substitution can't crack — like $\int x e^x\,dx$.
- The trick: trade a hard integral for an easier one by moving the derivative from one factor to the other.
- It's the BC workhorse for products of unlike functions.
把乘积法则倒着运行
- 链式法则反转给出换元积分。什么反转乘积法则?分部积分。
- 它处理换元搞不定的乘积积分——如 $\int x e^x\,dx$。
- 诀窍:通过把导数从一个因子移到另一个,把难积分换成更易的。
- 它是 BC 里处理不同类函数乘积的主力。
Integration by parts is the reverse of which rule? · 分部积分是哪个规则的逆运算？
It undoes the product rule. · 它撤销了乘积法则。
The formula
- From the product rule, integrated:
-
$$\int u\,dv = uv - \int v\,du$$
- You pick part of the integrand to be $u$ (which you'll differentiate) and the rest to be $dv$ (which you'll integrate).
- Then $du=u'\,dx$ and $v=\int dv$, and you assemble $uv-\int v\,du$.
公式
- 由乘积法则积分:
-
$$\int u\,dv = uv - \int v\,du$$
- 你选被积函数的一部分作 $u$(将求导),其余作 $dv$(将积分)。
- 然后 $du=u'\,dx$,$v=\int dv$,再拼成 $uv-\int v\,du$。
A product to integrate · 需要积分的乘积
y = a·e^{bx}
Integration by parts trades $\int x e^x$ for the easier $\int e^x$ — moving the derivative off the algebraic factor. · 分部积分用$\int x e^x$换取更容易的$\int e^x$——将导数从代数因子移开。
The integration by parts formula is $\int u\,dv=$ · 分部积分公式是$\int u\,dv=$
$\int u\,dv=uv-\int v\,du$ (mind the minus). · $\int u\,dv=uv-\int v\,du$（注意负号）。
Choosing $u$ and $dv$ wisely
- Aim to make the new integral $\int v\,du$ simpler than the original.
- A helpful priority for $u$ (LIATE): Logarithm, Inverse trig, Algebraic, Trig, Exponential — pick $u$ from earliest on this list.
- So in $\int x e^x\,dx$: $u=x$ (algebraic) and $dv=e^x\,dx$.
- A good choice collapses the problem; a bad choice makes it worse.
明智地选 $u$ 与 $dv$
- 目标是让新积分 $\int v\,du$ 比原来更简单。
- 选 $u$ 的一个有用优先级(LIATE):对数、反三角、代数、三角、指数——从这个表靠前的选 $u$。
- 所以在 $\int x e^x\,dx$ 中:$u=x$(代数),$dv=e^x\,dx$。
- 好的选择会把问题化简;坏的选择使它更糟。
For · 支持 $\int x e^x\,dx$, a good choice is $u=$ · 对于$\int x e^x\,dx$，一个好的选择是$u=$
By LIATE, the algebraic $x$ is $u$; $dv=e^x\,dx$. · 根据 LIATE 规则，代数 $x$ 是 $u$；$dv=e^x\,dx$。
A common mnemonic for choosing $u$ is ____ (L-I-A-T-E). · 选择$u$的常用助记符是____（L-I-A-T-E）。
Logarithm, Inverse trig, Algebraic, Trig, Exponential. · 对数、反三角、代数、三角、指数。
You should choose $u$ and $dv$ so the new integral $\int v\,du$ is simpler than the original. · 你应该选择$u$和$dv$使得新积分$\int v\,du$比原积分更简单。
That is the whole point of the choice. · 这正是选择的意义所在。
Sometimes repeat, sometimes loop
- If $\int v\,du$ is still a product, apply integration by parts again.
- For $\int x^2 e^x\,dx$ you'd use it twice, peeling one power of $x$ each time.
- A few integrals (like $\int e^x\sin x\,dx$) loop back to the original — solve algebraically for it.
- Keep $u,du,v,dv$ neatly labeled to avoid sign errors.
有时重复,有时循环
- 若 $\int v\,du$ 仍是乘积,就再次用分部积分。
- 对 $\int x^2 e^x\,dx$ 你要用两次,每次剥掉一个 $x$ 的幂。
- 少数积分(如 $\int e^x\sin x\,dx$)会循环回原来——用代数解出它。
- 把 $u,du,v,dv$ 整齐标注,以免符号出错。
$\displaystyle\int x e^x\,dx=$
$xe^x-\int e^x\,dx=xe^x-e^x+C$.
Choose $u$ and $dv$ so that $\int v\,du$ is simpler — the wrong pick can make it harder or send you in circles. Don't forget the minus sign: the formula is $uv-\int v\,du$. And $v$ is any antiderivative of $dv$ (drop its $+C$ until the very end of a definite/indefinite result).
选 $u$ 与 $dv$ 使 $\int v\,du$ 更简单——错误的选择会让它更难或让你兜圈子。别忘了减号:公式是 $uv-\int v\,du$。而 $v$ 是 $dv$ 的任意原函数(在最终结果之前丢掉它的 $+C$)。
Evaluate $\displaystyle\int x e^x\,dx$.
- Let $u=x$, $dv=e^x\,dx$. Then $du=dx$, $v=e^x$.
- $\displaystyle\int x e^x\,dx = uv-\int v\,du = x e^x - \int e^x\,dx$.
- $= x e^x - e^x + C = e^x(x-1)+C$.
求 $\displaystyle\int x e^x\,dx$。
- 令 $u=x$,$dv=e^x\,dx$。则 $du=dx$,$v=e^x$。
- $\displaystyle\int x e^x\,dx = uv-\int v\,du = x e^x - \int e^x\,dx$。
- $= x e^x - e^x + C = e^x(x-1)+C$。
Integration by parts reverses the product rule: $\int u\,dv=uv-\int v\,du$. Choose $u$ (to differentiate) and $dv$ (to integrate) — via LIATE — so that $\int v\,du$ is simpler. Mind the minus sign, and apply it repeatedly (or solve the loop) when needed.
分部积分反转乘积法则:$\int u\,dv=uv-\int v\,du$。经由 LIATE 选 $u$(求导)与 $dv$(积分),使 $\int v\,du$ 更简单。注意减号,必要时反复应用(或解循环)。