Unit B1: Integration Techniques I: Substitution and Parts单元 B1：积分技巧 I：换元法与分部积分

Evaluate $\displaystyle\int x\,e^{x^2}\,dx$. Let $u=x^2$, so $du=2x\,dx$ and $x\,dx=\tfrac12\,du$.

$$\int x\,e^{x^2}\,dx=\frac12\int e^{u}\,du=\frac12 e^{u}+C=\frac12 e^{x^2}+C.$$

The factor $x$ outside the exponential is exactly what the substitution consumes, up to the constant $\tfrac12$.

求 $\displaystyle\int x\,e^{x^2}\,dx$。令 $u=x^2$，则 $du=2x\,dx$，故 $x\,dx=\tfrac12\,du$。

$$\int x\,e^{x^2}\,dx=\frac12\int e^{u}\,du=\frac12 e^{u}+C=\frac12 e^{x^2}+C.$$

指数外的因子 $x$ 恰好被换元消耗，相差常数 $\tfrac12$。

Evaluate $\displaystyle\int \frac{\ln x}{x}\,dx$. Let $u=\ln x$, so $du=\tfrac{1}{x}\,dx$.

$$\int \frac{\ln x}{x}\,dx=\int u\,du=\frac{u^2}{2}+C=\frac{(\ln x)^2}{2}+C.$$

The deciding observation is that the factor $\tfrac{1}{x}$ is precisely $\tfrac{d}{dx}\ln x$. The whole integrand is therefore $u\,du$ in disguise, and the answer is a power of $u$. Differentiating $\tfrac12(\ln x)^2$ by the chain rule returns $(\ln x)\cdot\tfrac1x$, confirming the result.

求 $\displaystyle\int \frac{\ln x}{x}\,dx$。令 $u=\ln x$，则 $du=\tfrac{1}{x}\,dx$。

$$\int \frac{\ln x}{x}\,dx=\int u\,du=\frac{u^2}{2}+C=\frac{(\ln x)^2}{2}+C.$$

关键观察在于：因子 $\tfrac{1}{x}$ 恰好是 $\tfrac{d}{dx}\ln x$。因此整个被积函数本质上是 $u\,du$，答案是 $u$ 的幂次。对 $\tfrac12(\ln x)^2$ 用链式法则求导得 $(\ln x)\cdot\tfrac1x$，验证了结果。

Evaluate $\displaystyle\int \frac{x}{\sqrt{4-x^2}}\,dx$. Let $u=4-x^2$, so $du=-2x\,dx$, which gives $x\,dx=-\tfrac12\,du$. The integrand's $\sqrt{4-x^2}$ becomes $\sqrt{u}=u^{1/2}$:

$$\int \frac{x}{\sqrt{4-x^2}}\,dx=-\frac12\int u^{-1/2}\,du=-\frac12\cdot 2u^{1/2}+C=-\sqrt{4-x^2}+C.$$

Differentiating $-\sqrt{4-x^2}$ returns $\tfrac{x}{\sqrt{4-x^2}}$, which checks. Note that the bare $x$ in the numerator is essential: without it, the integral $\int \tfrac{dx}{\sqrt{4-x^2}}=\arcsin(x/2)+C$ is a different problem entirely, solved by a trigonometric substitution rather than this algebraic one.

求 $\displaystyle\int \frac{x}{\sqrt{4-x^2}}\,dx$。令 $u=4-x^2$，则 $du=-2x\,dx$，故 $x\,dx=-\tfrac12\,du$。被积函数中的 $\sqrt{4-x^2}$ 变为 $\sqrt{u}=u^{1/2}$：

$$\int \frac{x}{\sqrt{4-x^2}}\,dx=-\frac12\int u^{-1/2}\,du=-\frac12\cdot 2u^{1/2}+C=-\sqrt{4-x^2}+C.$$

对 $-\sqrt{4-x^2}$ 求导得 $\tfrac{x}{\sqrt{4-x^2}}$，验证正确。注意分子中单独出现的 $x$ 是必不可少的：若没有它，$\int \tfrac{dx}{\sqrt{4-x^2}}=\arcsin(x/2)+C$ 是完全不同的问题，需用三角换元法而非此代数换元法求解。

Evaluate $\displaystyle\int \tan^{2}x\,\sec^{2}x\,dx$. The factor $\sec^2 x$ is the derivative of $\tan x$, which signals $u=\tan x$, $du=\sec^2 x\,dx$:

$$\int \tan^{2}x\,\sec^{2}x\,dx=\int u^{2}\,du=\frac{u^3}{3}+C=\frac{\tan^{3}x}{3}+C.$$

The art is in reading the integrand for a function paired with its own derivative. Here $\tan x$ and $\sec^2 x$ form exactly such a pair, so the integral collapses to a power rule in $u$.

求 $\displaystyle\int \tan^{2}x\,\sec^{2}x\,dx$。因子 $\sec^2 x$ 是 $\tan x$ 的导数，提示令 $u=\tan x$，$du=\sec^2 x\,dx$：

$$\int \tan^{2}x\,\sec^{2}x\,dx=\int u^{2}\,du=\frac{u^3}{3}+C=\frac{\tan^{3}x}{3}+C.$$

核心在于识别被积函数中某个函数与其自身导数配对的结构。这里 $\tan x$ 与 $\sec^2 x$ 恰好构成这样一对，积分因此化为 $u$ 的幂次法则。

Evaluate $\displaystyle\int_0^{2} x\,e^{x^2}\,dx$ using the change-of-variables rule for definite integrals. Let $u=x^2$, $du=2x\,dx$, so $x\,dx=\tfrac12\,du$. Convert the limits: when $x=0$, $u=0$; when $x=2$, $u=4$. The entire problem is then carried out in $u$:

$$\int_0^{2} x\,e^{x^2}\,dx=\frac12\int_0^{4} e^{u}\,du=\frac12\bigl[e^{u}\bigr]_0^{4}=\frac{e^{4}-1}{2}.$$

Because the limits were converted to $u$-values, there is no back-substitution to $x$. The alternative, finding the indefinite antiderivative $\tfrac12 e^{x^2}$ and evaluating at $x=2$ and $x=0$, gives the same $\tfrac12(e^4-1)$, but converting the limits is cleaner and removes a step where errors creep in.

利用定积分的换元公式求 $\displaystyle\int_0^{2} x\,e^{x^2}\,dx$。令 $u=x^2$，$du=2x\,dx$，故 $x\,dx=\tfrac12\,du$。转换积分限：当 $x=0$ 时 $u=0$；当 $x=2$ 时 $u=4$。整个计算在 $u$ 变量下进行：

$$\int_0^{2} x\,e^{x^2}\,dx=\frac12\int_0^{4} e^{u}\,du=\frac12\bigl[e^{u}\bigr]_0^{4}=\frac{e^{4}-1}{2}.$$

因为积分限已转换为 $u$ 的值，无需将结果回代为 $x$。另一种方法是求不定原函数（antiderivative）$\tfrac12 e^{x^2}$，再代入 $x=2$ 和 $x=0$，结果相同，均为 $\tfrac12(e^4-1)$，但转换积分限更简洁，也少了一步容易出错的环节。

Let $F$ be an antiderivative of $f$, so $F'=f$. By the chain rule,

$$\frac{d}{dx}\,F\bigl(g(x)\bigr)=F'\bigl(g(x)\bigr)\,g'(x)=f\bigl(g(x)\bigr)\,g'(x).$$

Integrating both sides recovers $\int f(g(x))g'(x)\,dx=F(g(x))+C=\int f(u)\,du$ with $u=g(x)$. Thus the substitution rule is not a separate axiom; it is the antiderivative form of the chain rule.

For a definite integral the same identity, combined with the fundamental theorem of calculus, gives the change-of-variables rule with transformed limits. If $g$ is continuously differentiable on $[a,b]$ and $f$ is continuous on the range of $g$, then

$$\int_a^b f\bigl(g(x)\bigr)\,g'(x)\,dx=\int_{g(a)}^{g(b)} f(u)\,du.$$

The proof is one line: let $F'=f$, so by the chain rule $F\circ g$ is an antiderivative of $(f\circ g)\,g'$, and the fundamental theorem gives $\bigl[F(g(x))\bigr]_a^b=F(g(b))-F(g(a))=\bigl[F(u)\bigr]_{g(a)}^{g(b)}$. The practical payoff is that once the limits are converted to $u$-values there is no need to revert to $x$ at the end.

设 $F$ 是 $f$ 的原函数，即 $F'=f$。由链式法则（Chain Rule）：

$$\frac{d}{dx}\,F\bigl(g(x)\bigr)=F'\bigl(g(x)\bigr)\,g'(x)=f\bigl(g(x)\bigr)\,g'(x).$$

两边积分即得 $\int f(g(x))g'(x)\,dx=F(g(x))+C=\int f(u)\,du$（其中 $u=g(x)$）。因此换元法不是一条独立的公理，而是链式法则在求原函数层面的体现。

对于定积分，同一等式结合微积分基本定理（FTC），给出带换元积分限的变量替换公式。若 $g$ 在 $[a,b]$ 上连续可微，$f$ 在 $g$ 的值域上连续，则

$$\int_a^b f\bigl(g(x)\bigr)\,g'(x)\,dx=\int_{g(a)}^{g(b)} f(u)\,du.$$

证明只需一行：令 $F'=f$，由链式法则知 $F\circ g$ 是 $(f\circ g)\,g'$ 的原函数，再由微积分基本定理得 $\bigl[F(g(x))\bigr]_a^b=F(g(b))-F(g(a))=\bigl[F(u)\bigr]_{g(a)}^{g(b)}$。实际意义在于：一旦积分限转换为 $u$ 的值，最终无需回代为 $x$。

Evaluate $\displaystyle\int x\,e^{x}\,dx$. Take $u=x$ (algebraic, differentiates to a constant) and $dv=e^{x}\,dx$, so $du=dx$ and $v=e^{x}$.

$$\int x\,e^{x}\,dx=x e^{x}-\int e^{x}\,dx=x e^{x}-e^{x}+C=(x-1)e^{x}+C.$$

求 $\displaystyle\int x\,e^{x}\,dx$。令 $u=x$（代数型，微分后为常数），$dv=e^{x}\,dx$，则 $du=dx$，$v=e^{x}$。

$$\int x\,e^{x}\,dx=x e^{x}-\int e^{x}\,dx=x e^{x}-e^{x}+C=(x-1)e^{x}+C.$$

Evaluate $\displaystyle\int \ln x\,dx$. There is no obvious product, so take $u=\ln x$ and $dv=dx$, giving $du=\tfrac{1}{x}\,dx$ and $v=x$.

$$\int \ln x\,dx=x\ln x-\int x\cdot\frac{1}{x}\,dx=x\ln x-\int 1\,dx=x\ln x-x+C.$$

The same idea handles every isolated function whose derivative is simpler than itself. The hidden $dv=dx$ supplies $v=x$, and the integrand $v\,du$ becomes elementary.

求 $\displaystyle\int \ln x\,dx$。没有明显的乘积结构，令 $u=\ln x$，$dv=dx$，则 $du=\tfrac{1}{x}\,dx$，$v=x$。

$$\int \ln x\,dx=x\ln x-\int x\cdot\frac{1}{x}\,dx=x\ln x-\int 1\,dx=x\ln x-x+C.$$

同样的思路适用于所有"导数比自身更简单"的孤立函数。隐含的 $dv=dx$ 提供 $v=x$，被积量 $v\,du$ 随即变为初等形式。

Evaluate $\displaystyle\int \arctan x\,dx$. As with $\ln x$, take $u=\arctan x$ and $dv=dx$, so $du=\tfrac{1}{1+x^2}\,dx$ and $v=x$:

$$\int \arctan x\,dx=x\arctan x-\int \frac{x}{1+x^2}\,dx.$$

The residual integral is a substitution, $w=1+x^2$, $dw=2x\,dx$, giving $\int\tfrac{x}{1+x^2}\,dx=\tfrac12\ln(1+x^2)$. Therefore

$$\int \arctan x\,dx=x\arctan x-\frac12\ln(1+x^2)+C.$$

This is the canonical example of an integral that needs parts to start and substitution to finish. The Inverse-trig "I" in LIATE outranks everything else, which is exactly why $\arctan x$ becomes $u$.

求 $\displaystyle\int \arctan x\,dx$。与 $\ln x$ 同理，令 $u=\arctan x$，$dv=dx$，则 $du=\tfrac{1}{1+x^2}\,dx$，$v=x$：

$$\int \arctan x\,dx=x\arctan x-\int \frac{x}{1+x^2}\,dx.$$

剩余积分用换元法，令 $w=1+x^2$，$dw=2x\,dx$，得 $\int\tfrac{x}{1+x^2}\,dx=\tfrac12\ln(1+x^2)$。因此

$$\int \arctan x\,dx=x\arctan x-\frac12\ln(1+x^2)+C.$$

这是"分部积分开头、换元收尾"的典型例子。LIATE 中反三角函数（"I"）优先级最高，这正是 $\arctan x$ 成为 $u$ 的原因。

Evaluate $\displaystyle\int x\sin x\,dx$. By LIATE the algebraic factor $x$ becomes $u$ and $dv=\sin x\,dx$, so $du=dx$ and $v=-\cos x$:

$$\int x\sin x\,dx=-x\cos x-\int(-\cos x)\,dx=-x\cos x+\int\cos x\,dx=-x\cos x+\sin x+C.$$

The sign bookkeeping is where most slips occur: $v=-\cos x$ carries a minus that then meets the minus in $-\int v\,du$, producing the $+\int\cos x\,dx$. Differentiating $-x\cos x+\sin x$ returns $-\cos x+x\sin x+\cos x=x\sin x$, which confirms the answer.

求 $\displaystyle\int x\sin x\,dx$。按 LIATE，代数因子 $x$ 作为 $u$，$dv=\sin x\,dx$，则 $du=dx$，$v=-\cos x$：

$$\int x\sin x\,dx=-x\cos x-\int(-\cos x)\,dx=-x\cos x+\int\cos x\,dx=-x\cos x+\sin x+C.$$

符号追踪是最容易出错的地方：$v=-\cos x$ 带有一个负号，与 $-\int v\,du$ 中的负号相遇，产生 $+\int\cos x\,dx$。对 $-x\cos x+\sin x$ 求导得 $-\cos x+x\sin x+\cos x=x\sin x$，验证正确。

Let $u(x)$ and $v(x)$ be differentiable. The product rule gives

$$\frac{d}{dx}\bigl(u v\bigr)=u\,\frac{dv}{dx}+v\,\frac{du}{dx}.$$

Integrate both sides with respect to $x$. The left side integrates to $uv$, so

$$uv=\int u\,dv+\int v\,du \quad\Longrightarrow\quad \int u\,dv=uv-\int v\,du.$$

设 $u(x)$ 和 $v(x)$ 均可微。由乘积法则（Product Rule）：

$$\frac{d}{dx}\bigl(u v\bigr)=u\,\frac{dv}{dx}+v\,\frac{du}{dx}.$$

两边对 $x$ 积分，左边积分得 $uv$，因此

$$uv=\int u\,dv+\int v\,du \quad\Longrightarrow\quad \int u\,dv=uv-\int v\,du.$$

Evaluate $\displaystyle\int x^2 e^{x}\,dx$. First take $u=x^2$, $dv=e^{x}\,dx$:

$$\int x^2 e^{x}\,dx=x^2 e^{x}-2\int x e^{x}\,dx.$$

From Worked Example 2.1, $\int x e^{x}\,dx=(x-1)e^{x}+C$. Hence

$$\int x^2 e^{x}\,dx=x^2 e^{x}-2(x-1)e^{x}+C=(x^2-2x+2)e^{x}+C.$$

求 $\displaystyle\int x^2 e^{x}\,dx$。先令 $u=x^2$，$dv=e^{x}\,dx$：

$$\int x^2 e^{x}\,dx=x^2 e^{x}-2\int x e^{x}\,dx.$$

由例题 2.1，$\int x e^{x}\,dx=(x-1)e^{x}+C$。故

$$\int x^2 e^{x}\,dx=x^2 e^{x}-2(x-1)e^{x}+C=(x^2-2x+2)e^{x}+C.$$

Evaluate $\displaystyle I=\int e^{x}\sin x\,dx$. Take $u=e^{x}$, $dv=\sin x\,dx$, so $v=-\cos x$:

$$I=-e^{x}\cos x+\int e^{x}\cos x\,dx.$$

Apply parts again to the new integral with $u=e^{x}$, $dv=\cos x\,dx$, $v=\sin x$:

$$\int e^{x}\cos x\,dx=e^{x}\sin x-\int e^{x}\sin x\,dx=e^{x}\sin x-I.$$

Substitute and solve: $I=-e^{x}\cos x+e^{x}\sin x-I$, so $2I=e^{x}(\sin x-\cos x)$ and

$$I=\frac{e^{x}(\sin x-\cos x)}{2}+C.$$

The constant $C$ is reinstated only at the very end. While solving algebraically for $I$ we treat $I$ as a single unknown quantity and the bare antiderivatives without their constants; the family of antiderivatives is recovered by appending $+C$ once $I$ is isolated.

求 $\displaystyle I=\int e^{x}\sin x\,dx$。令 $u=e^{x}$，$dv=\sin x\,dx$，则 $v=-\cos x$：

$$I=-e^{x}\cos x+\int e^{x}\cos x\,dx.$$

对新积分再次应用分部积分，令 $u=e^{x}$，$dv=\cos x\,dx$，$v=\sin x$：

$$\int e^{x}\cos x\,dx=e^{x}\sin x-\int e^{x}\sin x\,dx=e^{x}\sin x-I.$$

代入并求解：$I=-e^{x}\cos x+e^{x}\sin x-I$，故 $2I=e^{x}(\sin x-\cos x)$，从而

$$I=\frac{e^{x}(\sin x-\cos x)}{2}+C.$$

积分常数 $C$ 仅在最后代数解出 $I$ 之后加回。求解过程中，将 $I$ 视为单一未知量，原函数不携带常数；求出 $I$ 后再补上 $+C$ 即得全体原函数。

Evaluate $\displaystyle I=\int e^{2x}\cos(3x)\,dx$, an instance of the boxed formula with $a=2$, $b=3$. Take $u=e^{2x}$, $dv=\cos(3x)\,dx$, so $v=\tfrac13\sin(3x)$:

$$I=\frac{1}{3}e^{2x}\sin(3x)-\frac{2}{3}\int e^{2x}\sin(3x)\,dx.$$

Apply parts again to the new integral with $u=e^{2x}$, $dv=\sin(3x)\,dx$, $v=-\tfrac13\cos(3x)$:

$$\int e^{2x}\sin(3x)\,dx=-\frac{1}{3}e^{2x}\cos(3x)+\frac{2}{3}\int e^{2x}\cos(3x)\,dx=-\frac{1}{3}e^{2x}\cos(3x)+\frac{2}{3}I.$$

Substitute back: $I=\tfrac13 e^{2x}\sin(3x)-\tfrac23\bigl(-\tfrac13 e^{2x}\cos(3x)+\tfrac23 I\bigr)$, which is $I=\tfrac13 e^{2x}\sin(3x)+\tfrac29 e^{2x}\cos(3x)-\tfrac49 I$. Collecting, $\tfrac{13}{9}I=\tfrac19 e^{2x}\bigl(3\sin(3x)+2\cos(3x)\bigr)$, so

$$I=\frac{e^{2x}\bigl(2\cos(3x)+3\sin(3x)\bigr)}{13}+C,$$

which matches the closed form $\tfrac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}$ with $a^2+b^2=4+9=13$. Memorizing the closed form is convenient, but reconstructing it by two passes of parts is a reliable fallback when the formula slips.

求 $\displaystyle I=\int e^{2x}\cos(3x)\,dx$，这是方框公式中 $a=2$、$b=3$ 的特例。令 $u=e^{2x}$，$dv=\cos(3x)\,dx$，则 $v=\tfrac13\sin(3x)$：

$$I=\frac{1}{3}e^{2x}\sin(3x)-\frac{2}{3}\int e^{2x}\sin(3x)\,dx.$$

对新积分再次应用分部积分，令 $u=e^{2x}$，$dv=\sin(3x)\,dx$，$v=-\tfrac13\cos(3x)$：

$$\int e^{2x}\sin(3x)\,dx=-\frac{1}{3}e^{2x}\cos(3x)+\frac{2}{3}\int e^{2x}\cos(3x)\,dx=-\frac{1}{3}e^{2x}\cos(3x)+\frac{2}{3}I.$$

代入：$I=\tfrac13 e^{2x}\sin(3x)-\tfrac23\bigl(-\tfrac13 e^{2x}\cos(3x)+\tfrac23 I\bigr)$，即 $I=\tfrac13 e^{2x}\sin(3x)+\tfrac29 e^{2x}\cos(3x)-\tfrac49 I$。整理得 $\tfrac{13}{9}I=\tfrac19 e^{2x}\bigl(3\sin(3x)+2\cos(3x)\bigr)$，故

$$I=\frac{e^{2x}\bigl(2\cos(3x)+3\sin(3x)\bigr)}{13}+C,$$

与封闭公式 $\tfrac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}$（$a^2+b^2=4+9=13$）吻合。记住封闭公式固然方便，但用两次分部积分重新推导是忘记公式时的可靠备用方法。

Evaluate $\displaystyle\int x^2 \cos x\,dx$. This is the polynomial-attrition pattern, not the cyclic one: each pass lowers the degree of $x^2$. Take $u=x^2$, $dv=\cos x\,dx$, $v=\sin x$:

$$\int x^2\cos x\,dx=x^2\sin x-2\int x\sin x\,dx.$$

From Worked Example 2.4, $\int x\sin x\,dx=-x\cos x+\sin x$. Substitute:

$$\int x^2\cos x\,dx=x^2\sin x-2\bigl(-x\cos x+\sin x\bigr)+C=x^2\sin x+2x\cos x-2\sin x+C.$$

Two applications suffice because the degree dropped from $2$ to $1$ to $0$. A differentiation check returns $2x\sin x+x^2\cos x+2\cos x-2x\sin x-2\cos x=x^2\cos x$, as required. Contrast this with Worked Example 3.2, where neither factor was a polynomial and the integral instead had to be closed algebraically.

求 $\displaystyle\int x^2 \cos x\,dx$。这是多项式消耗型而非循环型：每次次数降一。令 $u=x^2$，$dv=\cos x\,dx$，$v=\sin x$：

$$\int x^2\cos x\,dx=x^2\sin x-2\int x\sin x\,dx.$$

由例题 2.4，$\int x\sin x\,dx=-x\cos x+\sin x$。代入：

$$\int x^2\cos x\,dx=x^2\sin x-2\bigl(-x\cos x+\sin x\bigr)+C=x^2\sin x+2x\cos x-2\sin x+C.$$

两次应用即足，因为次数从 $2$ 降到 $1$ 再到 $0$。验证：对结果求导得 $2x\sin x+x^2\cos x+2\cos x-2x\sin x-2\cos x=x^2\cos x$，正确。对比例题 3.2——该例中两个因子都不是多项式，积分必须代数求解。

Evaluate $\displaystyle\int x^3 \sin x\,dx$. Differentiate $x^3$ in the left column and integrate $\sin x$ in the right column:

$$\begin{array}{c|cc}\text{sign}&D\;(x^3)&I\;(\sin x)\\\hline +&x^3&-\cos x\\ -&3x^2&-\sin x\\ +&6x&\cos x\\ -&6&\sin x\\ +&0&\end{array}$$

Multiply along the diagonals with the leading signs:

$$\int x^3\sin x\,dx=-x^3\cos x+3x^2\sin x+6x\cos x-6\sin x+C.$$

求 $\displaystyle\int x^3 \sin x\,dx$。左列对 $x^3$ 逐次微分，右列对 $\sin x$ 逐次积分：

$$\begin{array}{c|cc}\text{符号}&D\;(x^3)&I\;(\sin x)\\\hline +&x^3&-\cos x\\ -&3x^2&-\sin x\\ +&6x&\cos x\\ -&6&\sin x\\ +&0&\end{array}$$

沿对角线相乘并加上各项符号：

$$\int x^3\sin x\,dx=-x^3\cos x+3x^2\sin x+6x\cos x-6\sin x+C.$$

For $\displaystyle\int x^2 e^{2x}\,dx$ the left column terminates at $0$ after two derivatives, and antiderivatives of $e^{2x}$ are $\tfrac12 e^{2x},\tfrac14 e^{2x},\tfrac18 e^{2x}$. Diagonals give

$$\int x^2 e^{2x}\,dx=\frac{x^2}{2}e^{2x}-\frac{2x}{4}e^{2x}+\frac{2}{8}e^{2x}+C=\frac{e^{2x}}{4}\bigl(2x^2-2x+1\bigr)+C.$$

The care point here is that each step down the right column integrates the previous antiderivative, not the original $e^{2x}$. Because $\int e^{2x}\,dx=\tfrac12 e^{2x}$, every descent multiplies in another $\tfrac12$, producing $\tfrac12,\tfrac14,\tfrac18$. A common slip is to write $e^{2x},e^{2x},e^{2x}$ down the column, forgetting that integrating $e^{2x}$ repeatedly accumulates factors of $\tfrac12$.

对于 $\displaystyle\int x^2 e^{2x}\,dx$，左列经两次微分后归零，$e^{2x}$ 的逐次原函数为 $\tfrac12 e^{2x},\tfrac14 e^{2x},\tfrac18 e^{2x}$。沿对角线计算：

$$\int x^2 e^{2x}\,dx=\frac{x^2}{2}e^{2x}-\frac{2x}{4}e^{2x}+\frac{2}{8}e^{2x}+C=\frac{e^{2x}}{4}\bigl(2x^2-2x+1\bigr)+C.$$

注意：右列每向下一步，是对上一个原函数再次积分，而非对原始 $e^{2x}$ 积分。由于 $\int e^{2x}\,dx=\tfrac12 e^{2x}$，每次下降都累积一个 $\tfrac12$ 因子，得到 $\tfrac12,\tfrac14,\tfrac18$。常见错误是在右列全部写成 $e^{2x},e^{2x},e^{2x}$，忘记了重复积分会积累 $\tfrac12$ 因子。

Tabular integration also handles cases where the derivative column never reaches $0$, provided you stop and pick up a residual integral. Consider $\displaystyle\int x\,\sec^2 x\,dx$. Differentiate $x$ (reaching $0$ after one step) and integrate $\sec^2 x$ to $\tan x$:

$$\begin{array}{c|cc}\text{sign}&D\;(x)&I\;(\sec^2 x)\\\hline +&x&\tan x\\ -&1&\ldots\end{array}$$

The first diagonal gives $+x\tan x$. The next row pairs the derivative $1$ with the antiderivative $\tan x$ under the running sign $-$, leaving the residual $-\int 1\cdot\tan x\,dx$. Since $\int\tan x\,dx=-\ln|\cos x|$, we obtain

$$\int x\sec^2 x\,dx=x\tan x-\int \tan x\,dx=x\tan x+\ln|\cos x|+C.$$

Here the polynomial $x$ does terminate, so a single diagonal plus one residual integral finishes the job. The lesson is that tabular bookkeeping is valid even when the second column does not integrate to itself.

表格积分法也能处理微分列无法归零的情形，只需在合适处停止并附加剩余积分。考虑 $\displaystyle\int x\,\sec^2 x\,dx$。对 $x$ 微分（一步后归零），对 $\sec^2 x$ 积分得 $\tan x$：

$$\begin{array}{c|cc}\text{符号}&D\;(x)&I\;(\sec^2 x)\\\hline +&x&\tan x\\ -&1&\ldots\end{array}$$

第一条对角线给出 $+x\tan x$。下一行将导数 $1$ 与原函数 $\tan x$ 在符号 $-$ 下配对，留下剩余积分 $-\int 1\cdot\tan x\,dx$。由于 $\int\tan x\,dx=-\ln|\cos x|$，得到

$$\int x\sec^2 x\,dx=x\tan x-\int \tan x\,dx=x\tan x+\ln|\cos x|+C.$$

这里多项式 $x$ 确实终止，因此一条对角线加一个剩余积分即可完成。这说明即使右列不是自积分型，表格记账方法依然有效。

Evaluate $\displaystyle\int_0^{1} x e^{x}\,dx$. With $u=x$, $dv=e^{x}\,dx$, $du=dx$, $v=e^{x}$:

$$\int_0^{1} x e^{x}\,dx=\Bigl[x e^{x}\Bigr]_0^{1}-\int_0^{1} e^{x}\,dx=\bigl(1\cdot e\bigr)-\Bigl[e^{x}\Bigr]_0^{1}=e-(e-1)=1.$$

求 $\displaystyle\int_0^{1} x e^{x}\,dx$。令 $u=x$，$dv=e^{x}\,dx$，则 $du=dx$，$v=e^{x}$：

$$\int_0^{1} x e^{x}\,dx=\Bigl[x e^{x}\Bigr]_0^{1}-\int_0^{1} e^{x}\,dx=\bigl(1\cdot e\bigr)-\Bigl[e^{x}\Bigr]_0^{1}=e-(e-1)=1.$$

Evaluate $\displaystyle\int_1^{e} \ln x\,dx$. Using $\int\ln x\,dx=x\ln x-x$ from Worked Example 2.2:

$$\int_1^{e}\ln x\,dx=\Bigl[x\ln x-x\Bigr]_1^{e}=\bigl(e\cdot 1-e\bigr)-\bigl(1\cdot 0-1\bigr)=0-(-1)=1.$$

At the upper limit $x=e$ we used $\ln e=1$, and at the lower limit $x=1$ we used $\ln 1=0$. The area under $\ln x$ from $1$ to $e$ is therefore exactly one square unit, a clean result that is easy to sanity-check against the graph: $\ln x$ rises from $0$ to $1$ across an interval of width $e-1\approx 1.718$, so an area near $1$ is plausible.

求 $\displaystyle\int_1^{e} \ln x\,dx$。利用例题 2.2 中的 $\int\ln x\,dx=x\ln x-x$：

$$\int_1^{e}\ln x\,dx=\Bigl[x\ln x-x\Bigr]_1^{e}=\bigl(e\cdot 1-e\bigr)-\bigl(1\cdot 0-1\bigr)=0-(-1)=1.$$

在上限 $x=e$ 处用 $\ln e=1$，在下限 $x=1$ 处用 $\ln 1=0$。因此 $\ln x$ 在 $1$ 到 $e$ 的面积恰好是一个平方单位，这是一个优美的结果：$\ln x$ 在宽度为 $e-1\approx 1.718$ 的区间上从 $0$ 升至 $1$，面积约为 $1$ 是合理的。

Evaluate $\displaystyle\int_0^{1} \arctan x\,dx$. Take $u=\arctan x$, $dv=dx$, so $du=\tfrac{1}{1+x^2}\,dx$, $v=x$:

$$\int_0^{1}\arctan x\,dx=\Bigl[x\arctan x\Bigr]_0^{1}-\int_0^{1}\frac{x}{1+x^2}\,dx.$$

The boundary term is $1\cdot\arctan 1-0=\tfrac{\pi}{4}$. For the residual, substitute $w=1+x^2$, $dw=2x\,dx$; the limits become $w(0)=1$ and $w(1)=2$:

$$\int_0^{1}\frac{x}{1+x^2}\,dx=\frac12\int_1^{2}\frac{dw}{w}=\frac12\bigl[\ln w\bigr]_1^{2}=\frac12\ln 2.$$

Therefore $\displaystyle\int_0^{1}\arctan x\,dx=\frac{\pi}{4}-\frac12\ln 2\approx 0.4388$. The boundary term stayed in $x$ at the original endpoints, while the residual integral switched cleanly to $w$ with its own limits $1$ and $2$. This is the disciplined way to combine the two methods.

求 $\displaystyle\int_0^{1} \arctan x\,dx$。令 $u=\arctan x$，$dv=dx$，则 $du=\tfrac{1}{1+x^2}\,dx$，$v=x$：

$$\int_0^{1}\arctan x\,dx=\Bigl[x\arctan x\Bigr]_0^{1}-\int_0^{1}\frac{x}{1+x^2}\,dx.$$

边界项为 $1\cdot\arctan 1-0=\tfrac{\pi}{4}$。对剩余积分，令 $w=1+x^2$，$dw=2x\,dx$；积分限变为 $w(0)=1$，$w(1)=2$：

$$\int_0^{1}\frac{x}{1+x^2}\,dx=\frac12\int_1^{2}\frac{dw}{w}=\frac12\bigl[\ln w\bigr]_1^{2}=\frac12\ln 2.$$

因此 $\displaystyle\int_0^{1}\arctan x\,dx=\frac{\pi}{4}-\frac12\ln 2\approx 0.4388$。边界项始终在原变量 $x$ 下于原端点处求值，而剩余积分干净地切换至 $w$，使用各自的积分限 $1$ 和 $2$。这是规范结合两种方法的做法。

Evaluate $\displaystyle\int \cos(\sqrt{x})\,dx$. Substitute $t=\sqrt{x}$, so $x=t^2$ and $dx=2t\,dt$:

$$\int\cos(\sqrt{x})\,dx=\int \cos t\,(2t)\,dt=2\int t\cos t\,dt.$$

Now use parts with $u=t$, $dv=\cos t\,dt$: $\int t\cos t\,dt=t\sin t+\cos t$. Therefore

$$\int\cos(\sqrt{x})\,dx=2\bigl(\sqrt{x}\sin\sqrt{x}+\cos\sqrt{x}\bigr)+C.$$

求 $\displaystyle\int \cos(\sqrt{x})\,dx$。令 $t=\sqrt{x}$，则 $x=t^2$，$dx=2t\,dt$：

$$\int\cos(\sqrt{x})\,dx=\int \cos t\,(2t)\,dt=2\int t\cos t\,dt.$$

再用分部积分，令 $u=t$，$dv=\cos t\,dt$：$\int t\cos t\,dt=t\sin t+\cos t$。因此

$$\int\cos(\sqrt{x})\,dx=2\bigl(\sqrt{x}\sin\sqrt{x}+\cos\sqrt{x}\bigr)+C.$$

Evaluate $\displaystyle\int x^3\,e^{x^2}\,dx$. This looks like parts, but write $x^3=x^2\cdot x$ and substitute $u=x^2$, $du=2x\,dx$:

$$\int x^3 e^{x^2}\,dx=\frac12\int u\,e^{u}\,du=\frac12\bigl(u-1\bigr)e^{u}+C=\frac12\bigl(x^2-1\bigr)e^{x^2}+C,$$

where the inner $\int u e^{u}\,du$ is finished by parts. Substitution first removed the awkward $x^2$ in the exponent.

求 $\displaystyle\int x^3\,e^{x^2}\,dx$。看似分部积分，但将 $x^3=x^2\cdot x$，令 $u=x^2$，$du=2x\,dx$：

$$\int x^3 e^{x^2}\,dx=\frac12\int u\,e^{u}\,du=\frac12\bigl(u-1\bigr)e^{u}+C=\frac12\bigl(x^2-1\bigr)e^{x^2}+C,$$

其中内层 $\int u e^{u}\,du$ 用分部积分收尾。先换元消去了指数中麻烦的 $x^2$。

Evaluate $\displaystyle\int \frac{x^2}{x^2+1}\,dx$. Neither substitution nor parts attacks this directly, but algebraic division does: $\tfrac{x^2}{x^2+1}=1-\tfrac{1}{x^2+1}$. Now each piece is standard:

$$\int \frac{x^2}{x^2+1}\,dx=\int 1\,dx-\int \frac{1}{x^2+1}\,dx=x-\arctan x+C.$$

The lesson generalizes: when the degree of the numerator is at least the degree of the denominator, divide first. The leftover proper fraction is then a substitution, a partial-fraction problem, or a recognizable arctangent form.

求 $\displaystyle\int \frac{x^2}{x^2+1}\,dx$。换元法和分部积分法都无法直接处理，但多项式除法可以：$\tfrac{x^2}{x^2+1}=1-\tfrac{1}{x^2+1}$。每一项都是标准形式：

$$\int \frac{x^2}{x^2+1}\,dx=\int 1\,dx-\int \frac{1}{x^2+1}\,dx=x-\arctan x+C.$$

这个方法可以推广：当分子次数不低于分母次数时，先做多项式除法。剩余的真分式再通过换元法、部分分式分解，或可识别的反正切形式处理。

Evaluate $\displaystyle\int \sin^2 x\,dx$. Parts cycles back on itself here, but the power-reduction identity $\sin^2 x=\tfrac{1-\cos 2x}{2}$ flattens the problem:

$$\int \sin^2 x\,dx=\int \frac{1-\cos 2x}{2}\,dx=\frac{x}{2}-\frac{\sin 2x}{4}+C.$$

The $\int\cos 2x\,dx=\tfrac12\sin 2x$ is itself a linear substitution. Recognizing that an identity, not a technique, was the right first move is exactly the judgment this section is meant to build.

求 $\displaystyle\int \sin^2 x\,dx$。分部积分在这里会循环，但降幂公式 $\sin^2 x=\tfrac{1-\cos 2x}{2}$ 可以将问题展开：

$$\int \sin^2 x\,dx=\int \frac{1-\cos 2x}{2}\,dx=\frac{x}{2}-\frac{\sin 2x}{4}+C.$$

其中 $\int\cos 2x\,dx=\tfrac12\sin 2x$ 本身是一个线性换元。认识到应先用恒等式而非技巧，正是本节要培养的判断力。

Use the boxed reduction formula to evaluate $\displaystyle\int \sin^4 x\,dx$. With $n=4$:

$$\int \sin^4 x\,dx=-\frac{\sin^3 x\cos x}{4}+\frac{3}{4}\int \sin^2 x\,dx.$$

Now apply it again with $n=2$, using $\int \sin^0 x\,dx=\int 1\,dx=x$:

$$\int \sin^2 x\,dx=-\frac{\sin x\cos x}{2}+\frac{1}{2}x.$$

Substitute back to assemble the final answer:

$$\int \sin^4 x\,dx=-\frac{\sin^3 x\cos x}{4}-\frac{3\sin x\cos x}{8}+\frac{3x}{8}+C.$$

The reduction formula turned a fourth-power integral into a finite descent: $n=4\to n=2\to n=0$, two applications and a trivial base case. Without it, one would face several nested rounds of parts or a sequence of double-angle identities.

用方框中的递推公式求 $\displaystyle\int \sin^4 x\,dx$。取 $n=4$：

$$\int \sin^4 x\,dx=-\frac{\sin^3 x\cos x}{4}+\frac{3}{4}\int \sin^2 x\,dx.$$

再用 $n=2$ 应用一次，利用基础情形 $\int \sin^0 x\,dx=\int 1\,dx=x$：

$$\int \sin^2 x\,dx=-\frac{\sin x\cos x}{2}+\frac{1}{2}x.$$

回代得最终答案：

$$\int \sin^4 x\,dx=-\frac{\sin^3 x\cos x}{4}-\frac{3\sin x\cos x}{8}+\frac{3x}{8}+C.$$

递推公式将四次幂积分化为有限下降：$n=4\to n=2\to n=0$，两次应用加一个平凡的基础情形。若没有它，需要面对多轮嵌套的分部积分或一系列二倍角恒等式。

Reduction formulas are not limited to trigonometric powers. A single application of parts to $\int x^n e^{x}\,dx$ with $u=x^n$, $dv=e^{x}\,dx$ gives

$$\int x^n e^{x}\,dx=x^n e^{x}-n\int x^{n-1}e^{x}\,dx.$$

The recursion drops the exponent by one each time, terminating at $\int x^0 e^{x}\,dx=e^{x}$. Apply it to $\int x^3 e^{x}\,dx$: with $n=3$, then $n=2$, then $n=1$,

$$\int x^3 e^{x}\,dx=x^3 e^{x}-3\Bigl(x^2 e^{x}-2\bigl(x e^{x}-e^{x}\bigr)\Bigr)+C=\bigl(x^3-3x^2+6x-6\bigr)e^{x}+C.$$

This is the same answer the tabular method of Section 4 produces, which is no coincidence: tabular integration is precisely this reduction recursion written as a table. The reduction formula makes the underlying recursion explicit and easy to prove by induction.

递推公式不仅适用于三角函数幂次。对 $\int x^n e^{x}\,dx$ 做一次分部积分，令 $u=x^n$，$dv=e^{x}\,dx$，得

$$\int x^n e^{x}\,dx=x^n e^{x}-n\int x^{n-1}e^{x}\,dx.$$

每次递推指数降一，在 $\int x^0 e^{x}\,dx=e^{x}$ 处终止。对 $\int x^3 e^{x}\,dx$ 应用（$n=3$、$n=2$、$n=1$）：

$$\int x^3 e^{x}\,dx=x^3 e^{x}-3\Bigl(x^2 e^{x}-2\bigl(x e^{x}-e^{x}\bigr)\Bigr)+C=\bigl(x^3-3x^2+6x-6\bigr)e^{x}+C.$$

这与第四节表格法的结果相同，绝非巧合：表格积分法正是将此递推写成表格形式。递推公式使底层递推显式化，易于用数学归纳法证明。

Write $\sin^n x=\sin^{n-1}x\cdot\sin x$ and apply parts with $u=\sin^{n-1}x$, $dv=\sin x\,dx$, so $du=(n-1)\sin^{n-2}x\cos x\,dx$ and $v=-\cos x$:

$$\int\sin^n x\,dx=-\sin^{n-1}x\cos x+(n-1)\int \sin^{n-2}x\cos^2 x\,dx.$$

Replace $\cos^2 x=1-\sin^2 x$:

$$\int\sin^n x\,dx=-\sin^{n-1}x\cos x+(n-1)\int\sin^{n-2}x\,dx-(n-1)\int\sin^{n}x\,dx.$$

Move the last term to the left and divide by $n$ to obtain the reduction formula stated above.

将 $\sin^n x=\sin^{n-1}x\cdot\sin x$，应用分部积分，令 $u=\sin^{n-1}x$，$dv=\sin x\,dx$，则 $du=(n-1)\sin^{n-2}x\cos x\,dx$，$v=-\cos x$：

$$\int\sin^n x\,dx=-\sin^{n-1}x\cos x+(n-1)\int \sin^{n-2}x\cos^2 x\,dx.$$

将 $\cos^2 x=1-\sin^2 x$ 代入：

$$\int\sin^n x\,dx=-\sin^{n-1}x\cos x+(n-1)\int\sin^{n-2}x\,dx-(n-1)\int\sin^{n}x\,dx.$$

将最后一项移到左侧并除以 $n$，即得上方方框中的递推公式。

Starting from the fundamental theorem $f(x)=f(a)+\int_a^{x} f'(t)\,dt$ and applying parts repeatedly with $dv=dt$ (choosing the antiderivative $v=-(x-t)$ so the boundary terms telescope) produces

$$f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k+\frac{1}{n!}\int_a^{x}(x-t)^{n}f^{(n+1)}(t)\,dt.$$

The integral on the right is the exact remainder. Thus repeated integration by parts is the engine behind Taylor's theorem with the integral form of the remainder.

It is worth seeing the first step in full, because it shows where the unusual antiderivative $v=-(x-t)$ comes from. In $\int_a^{x} f'(t)\,dt$ regard $t$ as the variable of integration with $x$ fixed, and apply parts with $u=f'(t)$ and $dv=dt$. The clever move is to take $v=-(x-t)$ rather than the obvious $v=t$; both are antiderivatives of $dt$ in $t$ (since $\tfrac{d}{dt}[-(x-t)]=1$), but the shifted choice makes the boundary term vanish at the upper limit. Then $du=f''(t)\,dt$ and

$$\int_a^{x} f'(t)\,dt=\Bigl[-(x-t)f'(t)\Bigr]_a^{x}+\int_a^{x}(x-t)f''(t)\,dt=(x-a)f'(a)+\int_a^{x}(x-t)f''(t)\,dt,$$

because the boundary term is $0$ at $t=x$ and $(x-a)f'(a)$ at $t=a$. This recovers the degree-one Taylor polynomial plus a remainder of the same shape with one higher derivative. Choosing $v=-\tfrac{(x-t)^2}{2}$ on the next pass yields the quadratic term, and in general $v=-\tfrac{(x-t)^{k}}{k}$ at the $k$-th step. Each application advances the polynomial by one term and raises the power of $(x-t)$ and the order of the derivative inside the remainder by one, giving the boxed formula by induction. The factorial $\tfrac{1}{n!}$ accumulates from the constants generated at each step.

从微积分基本定理（FTC）出发：$f(x)=f(a)+\int_a^{x} f'(t)\,dt$，反复应用分部积分（令 $dv=dt$，选取原函数 $v=-(x-t)$ 使边界项可相消），得到

$$f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k+\frac{1}{n!}\int_a^{x}(x-t)^{n}f^{(n+1)}(t)\,dt.$$

右边的积分是精确余项。因此多次分部积分是带积分余项形式的泰勒定理的驱动引擎。

第一步值得完整展开，因为它揭示了不寻常的原函数 $v=-(x-t)$ 的来源。在 $\int_a^{x} f'(t)\,dt$ 中，将 $t$ 视为积分变量（$x$ 固定），令 $u=f'(t)$，$dv=dt$，应用分部积分。巧妙之处在于选取 $v=-(x-t)$ 而非显然的 $v=t$；两者都是关于 $t$ 的 $dt$ 的原函数（因为 $\tfrac{d}{dt}[-(x-t)]=1$），但移位选择使边界项在上限处消失。于是 $du=f''(t)\,dt$，且

$$\int_a^{x} f'(t)\,dt=\Bigl[-(x-t)f'(t)\Bigr]_a^{x}+\int_a^{x}(x-t)f''(t)\,dt=(x-a)f'(a)+\int_a^{x}(x-t)f''(t)\,dt,$$

因为边界项在 $t=x$ 处为 $0$，在 $t=a$ 处为 $(x-a)f'(a)$。这恢复了一次泰勒多项式加上同形状但导数阶数更高一阶的余项。下一步选 $v=-\tfrac{(x-t)^2}{2}$ 给出二次项，一般地，第 $k$ 步选 $v=-\tfrac{(x-t)^{k}}{k}$。每次应用使多项式向前推进一项，并将余项中 $(x-t)$ 的幂次和导数阶数各提高一，由归纳法得到方框中的公式。阶乘 $\tfrac{1}{n!}$ 由每步产生的常数积累而来。