Integrating Functions Using Long Division and Completing the Square · 使用长除法和配方法积分函数
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| completing the square/kəmˈpliːtɪŋ ðə skweə/ | 配方 | pèi fāng |
| polynomial long division/ˌpɒlɪˈnəʊmɪəl lɒŋ dɪˈvɪʒn/ | 多项式长除法 | duō xiàng shì zhǎng chú fǎ |
Rewrite first, then integrate
- Some rational functions don't match any basic antiderivative — until you rewrite them.
- Two algebra tools prepare a stubborn integrand: long division and completing the square.
- Neither is calculus; both just reshape the fraction into a friendlier form.
- Then a basic rule or a known integral finishes the job.
先改写,再积分
- 有些有理函数不匹配任何基本原函数——直到你把它们改写。
- 两样代数工具能预处理顽固的被积函数:长除法和配方。
- 两者都不是微积分;它们只是把分式重塑成更友好的形式。
- 然后一条基本规则或一个已知积分就完成任务。
Long division for "top-heavy" fractions
- If the numerator's degree is $\ge$ the denominator's, the fraction is improper.
- Use polynomial long division 多项式长除法 to split it into a polynomial plus a proper remainder.
- $\dfrac{x^2}{x+1}=x-1+\dfrac{1}{x+1}$ — now each piece integrates easily.
- $\displaystyle\int\dfrac{x^2}{x+1}\,dx=\dfrac{x^2}{2}-x+\ln|x+1|+C$.
长除法处理"头重"分式
- 若分子的次数 $\ge$ 分母的,分式就是假分式。
- 用多项式长除法把它拆成一个多项式加一个真余式。
- $\dfrac{x^2}{x+1}=x-1+\dfrac{1}{x+1}$——现在每块都容易积分。
- $\displaystyle\int\dfrac{x^2}{x+1}\,dx=\dfrac{x^2}{2}-x+\ln|x+1|+C$。
A rational curve to rewrite · 待改写的不定分式曲线
y = a / (x − b)
An improper fraction like $\tfrac{x^2}{x+1}$ becomes a line plus a small remainder — each piece integrates easily. · 像$\tfrac{x^2}{x+1}$这样的不定分式可化为一条直线加上一个小余式——每一部分都容易积分。
Use polynomial long division when the numerator's degree is... · 当分子的次数...
An improper fraction (top-heavy) calls for division. · 不定分式（分子次数高）需要除法。
Long division gives $\dfrac{x^2}{x+1}=$ · 长除法得到$\dfrac{x^2}{x+1}=$
$x^2=(x+1)(x-1)+1$, so the quotient is $x-1$ remainder $1$. · $x^2=(x+1)(x-1)+1$，因此商式为$x-1$，余式为$1$。
Completing the square for inverse-trig forms
- A denominator like $x^2+4x+5$ hides a perfect square: complete the square 配方 to $(x+2)^2+1$.
- That $\dfrac{1}{(x+2)^2+1}$ shape matches the arctangent antiderivative.
- $\displaystyle\int\dfrac{1}{x^2+4x+5}\,dx=\int\dfrac{1}{(x+2)^2+1}\,dx=\arctan(x+2)+C$.
- Completing the square turns a messy quadratic denominator into the clean $u^2+a^2$ pattern.
配方以得到反三角形式
- 像 $x^2+4x+5$ 这样的分母藏着一个完全平方:配方成 $(x+2)^2+1$。
- 那个 $\dfrac{1}{(x+2)^2+1}$ 的形状匹配反正切原函数。
- $\displaystyle\int\dfrac{1}{x^2+4x+5}\,dx=\int\dfrac{1}{(x+2)^2+1}\,dx=\arctan(x+2)+C$。
- 配方把杂乱的二次分母变成干净的 $u^2+a^2$ 模式。
Completing the square, $x^2+4x+5=$ · 配方法，$x^2+4x+5=$
$(x+2)^2=x^2+4x+4$, so $x^2+4x+5=(x+2)^2+1$. · $(x+2)^2=x^2+4x+4$，所以 $x^2+4x+5=(x+2)^2+1$。
The form $\dfrac{1}{(x+2)^2+1}$ integrates to an ____ function. · 形式$\dfrac{1}{(x+2)^2+1}$的积分结果为____函数。
$\int\tfrac{1}{u^2+1}\,du=\arctan u+C$.
Recognize the signal
- Numerator degree $\ge$ denominator degree → reach for long division first.
- An irreducible quadratic denominator (no real roots) → try completing the square toward arctan.
- These are setup moves; the actual integration is a basic rule or u-sub afterward.
- Spotting which rewrite to use is half the battle.
认出信号
- 分子次数 $\ge$ 分母次数 → 先用长除法。
- 不可约的二次分母(无实根)→ 试配方朝反正切走。
- 这些是建立步骤;真正的积分随后是一条基本规则或换元。
- 看出该用哪种改写就成功了一半。
$\displaystyle\int\dfrac{x^2}{x+1}\,dx=$
Divide first, then integrate $x-1+\tfrac1{x+1}$. · 先除后积$x-1+\tfrac1{x+1}$。
Match the signal to the rewrite: select all · 所有 true pairings. · 将信号与改写匹配：选择所有正确的配对。
Long division is for improper fractions only; both are pre-integration rewrites. · 长除法仅用于不定分式；两者均为积分前的改写。
Do the long division only when the fraction is improper (numerator degree $\ge$ denominator's) — dividing a proper fraction gets you nowhere. And after completing the square, keep the constant right: $x^2+4x+5=(x+2)^2+1$, not $(x+2)^2+5$. A wrong constant breaks the arctan match.
只在分式是假分式时(分子次数 $\ge$ 分母的)才做长除法——对真分式做除法无济于事。而配方后,常数要对:$x^2+4x+5=(x+2)^2+1$,不是 $(x+2)^2+5$。常数错了就破坏反正切匹配。
Integrate $\displaystyle\int\dfrac{x^2}{x+1}\,dx$.
- Long division: $\dfrac{x^2}{x+1}=x-1+\dfrac{1}{x+1}$.
- Integrate term by term: $\dfrac{x^2}{2}-x+\ln|x+1|+C$.
- (Each rewritten piece uses a basic antiderivative rule.)
求 $\displaystyle\int\dfrac{x^2}{x+1}\,dx$。
- 长除法:$\dfrac{x^2}{x+1}=x-1+\dfrac{1}{x+1}$。
- 逐项积分:$\dfrac{x^2}{2}-x+\ln|x+1|+C$。
- (每个改写后的块用一条基本原函数规则。)
When a rational integrand fits no basic rule, rewrite it first: use polynomial long division on an improper fraction (numerator degree $\ge$ denominator's), or complete the square in an irreducible quadratic denominator to reach an arctangent form. Then finish with a basic antiderivative.
当有理被积函数不匹配任何基本规则时,先改写它:对假分式(分子次数 $\ge$ 分母的)用多项式长除法,或对不可约二次分母配方以得到反正切形式。然后用基本原函数收尾。