Integrating Using Substitution · 使用换元法积分
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| u-substitution/juː ˌsʌbstɪˈtjuːʃn/ | 换元法 | huàn yuán fǎ |
The chain rule, run in reverse
- The chain rule made derivatives of composite functions. Integration must undo it.
- u-substitution 换元法 (or u-sub) is the reverse of the chain rule.
- It spots a composite function together with its inner derivative and simplifies by renaming.
- Master it and a huge class of otherwise-impossible integrals becomes routine.
把链式法则倒着运行
- 链式法则给出复合函数的导数。积分必须撤销它。
- 换元积分(u-代换)是链式法则的逆过程。
- 它发现一个复合函数连同它的内层导数,通过改名来化简。
- 掌握它,一大类原本无从下手的积分就变成例行公事。
u-substitution is the reverse of which differentiation rule? · u代换是哪个求导法则的逆运算？
It undoes the chain rule. · 它抵消了链式法则。
Rename the inner function as $u$
- Look for an inner function $g(x)$ whose derivative $g'(x)$ also appears (up to a constant) in the integrand.
- Let $u=g(x)$; then $du=g'(x)\,dx$.
- Rewrite the whole integral in terms of $u$ — the messy $x$-expression turns into a simple $u$-integral.
- Then integrate in $u$ and substitute back $u=g(x)$ at the end.
把内层函数改名为 $u$
- 寻找一个内层函数 $g(x)$,其导数 $g'(x)$ 也(相差一个常数地)出现在被积函数中。
- 令 $u=g(x)$;则 $du=g'(x)\,dx$。
- 把整个积分改写成 $u$ 的形式——杂乱的 $x$ 表达式变成简单的 $u$ 积分。
- 然后对 $u$ 积分,最后代回 $u=g(x)$。
A composite ready for u-sub · 适合u代换的复合函数
y = ax³ + bx
An integrand like $2x(x^2+1)^3$ hides an inner function and its derivative — the signal to substitute. · 被积函数$2x(x^2+1)^3$中隐藏着内层函数及其导数——这是进行代换的信号。
For · 支持 $\int 2x(x^2+1)^3\,dx$, a good choice is $u=$ · 对于$\int 2x(x^2+1)^3\,dx$，一个好的选择是$u=$
$u=x^2+1$ has $du=2x\,dx$, which is present. · $u=x^2+1$包含$du=2x\,dx$，且该部分存在。
A clean example of the pattern
- $\displaystyle\int 2x\,(x^2+1)^3\,dx$. Inner function $u=x^2+1$, so $du=2x\,dx$.
- The $2x\,dx$ in the integral is exactly $du$ — a perfect match.
- The integral becomes $\displaystyle\int u^3\,du=\dfrac{u^4}{4}+C$.
- Substitute back: $\dfrac{(x^2+1)^4}{4}+C$.
这个模式的干净例子
- $\displaystyle\int 2x\,(x^2+1)^3\,dx$。内层函数 $u=x^2+1$,所以 $du=2x\,dx$。
- 积分里的 $2x\,dx$ 恰好是 $du$——完美匹配。
- 积分变成 $\displaystyle\int u^3\,du=\dfrac{u^4}{4}+C$。
- 代回:$\dfrac{(x^2+1)^4}{4}+C$。
$\int 2x(x^2+1)^3\,dx$ equals... · $\int 2x(x^2+1)^3\,dx$等于...
$\int u^3\,du=\tfrac{u^4}{4}+C$, then back-substitute. · $\int u^3\,du=\tfrac{u^4}{4}+C$，然后回代。
Handle the constant, or the limits
- If $du$ is off by a constant factor, just multiply/divide to fix it (e.g. $x\,dx=\tfrac12\,du$).
- For a definite integral, either substitute back to $x$ before evaluating, or change the limits to $u$-values.
- Changing limits is cleaner: convert $a,b$ into $u(a),u(b)$ and never return to $x$.
- Always end an indefinite answer in terms of the original variable $x$.
处理常数,或积分限
- 若 $du$ 相差一个常数因子,乘/除来修正即可(如 $x\,dx=\tfrac12\,du$)。
- 对定积分,要么在求值前代回 $x$,要么把积分限换成 $u$ 值。
- 换限更干净:把 $a,b$ 换成 $u(a),u(b)$,再也不回到 $x$。
- 不定积分的答案永远以原变量 $x$ 结束。
Evaluate · 评价 $\int_0^1 3x^2(x^3+1)^2\,dx$ (let $u=x^3+1$). Enter a decimal. · 计算$\int_0^1 3x^2(x^3+1)^2\,dx$（令$u=x^3+1$）。输入小数。
$\int_1^2 u^2\,du=\tfrac{8-1}{3}=\tfrac73\approx2.333$.
A correct u-substitution converts $dx$ entirely into $du$, leaving no stray $x$. · 正确的u代换将$dx$完全转换为$du$，不留任何残留的$x$。
A leftover $x$ or · 或 $dx$ is the classic mistake. · 残留的$x$或$dx$是经典错误。
u-substitution works when the integrand contains the inner function's... · 当被积函数包含内层函数的...时，u代换有效
The inner derivative $g'(x)$ must be present (times a constant). · 内层导数$g'(x)$必须存在（乘以常数即可）。
u-sub works only when the inner derivative is present (up to a constant). If $g'(x)$ isn't there, you can't just invent it. And don't forget to convert $dx$ into $du$ — leaving a stray $dx$ or a leftover $x$ is the classic error. For definite integrals, either change the limits or back-substitute, not a mix.
换元积分只在内层导数存在(相差一个常数)时有效。若 $g'(x)$ 不在,你不能凭空造出它。而且别忘了把 $dx$ 换成 $du$——留下一个多余的 $dx$ 或剩下的 $x$ 是经典错误。对定积分,要么换限要么代回,别混用。
Evaluate $\displaystyle\int_0^1 3x^2\,(x^3+1)^2\,dx$.
- Let $u=x^3+1$, so $du=3x^2\,dx$. Limits: $x=0\Rightarrow u=1$, $x=1\Rightarrow u=2$.
- Integral becomes $\displaystyle\int_1^2 u^2\,du=\Big[\tfrac{u^3}{3}\Big]_1^2=\tfrac{8}{3}-\tfrac{1}{3}=\tfrac{7}{3}$.
求 $\displaystyle\int_0^1 3x^2\,(x^3+1)^2\,dx$。
- 令 $u=x^3+1$,所以 $du=3x^2\,dx$。积分限:$x=0\Rightarrow u=1$,$x=1\Rightarrow u=2$。
- 积分变成 $\displaystyle\int_1^2 u^2\,du=\Big[\tfrac{u^3}{3}\Big]_1^2=\tfrac{8}{3}-\tfrac{1}{3}=\tfrac{7}{3}$。
u-substitution reverses the chain rule: set $u=g(x)$ (the inner function), replace $g'(x)\,dx$ with $du$, integrate in $u$, then substitute back (or change the limits for a definite integral). It works when the inner derivative appears up to a constant factor. Always convert $dx\to du$ fully.
换元积分反转链式法则:令 $u=g(x)$(内层函数),把 $g'(x)\,dx$ 换成 $du$,对 $u$ 积分,再代回(定积分则换限)。当内层导数相差一个常数因子地出现时它有效。永远把 $dx\to du$ 完全转换。