Unit B2: Integration Techniques II: Trig and Partial Fractions第 B2 单元：积分技巧 II：三角与部分分式

Evaluate $\int \sin^3 x \cos^2 x\,dx$. Sine has the odd power, so reserve one factor of $\sin x$ and convert the remaining $\sin^2 x = 1 - \cos^2 x$.

$$\int \sin^3 x \cos^2 x\,dx = \int (1-\cos^2 x)\cos^2 x \,\sin x\,dx.$$

Let $u = \cos x$, so $du = -\sin x\,dx$. Then

$$-\int (1-u^2)u^2\,du = -\int (u^2 - u^4)\,du = -\frac{u^3}{3} + \frac{u^5}{5} + C.$$ $$= \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C.$$

计算 $\int \sin^3 x \cos^2 x\,dx$。正弦是奇数次幂，故保留一个 $\sin x$ 因子，将剩余的 $\sin^2 x = 1 - \cos^2 x$ 转化。

$$\int \sin^3 x \cos^2 x\,dx = \int (1-\cos^2 x)\cos^2 x \,\sin x\,dx.$$

令 $u = \cos x$，则 $du = -\sin x\,dx$，代入得

$$-\int (1-u^2)u^2\,du = -\int (u^2 - u^4)\,du = -\frac{u^3}{3} + \frac{u^5}{5} + C.$$ $$= \frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C.$$

Evaluate $\int \cos^2 x\,dx$ using the half-angle identity.

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

For $\int \sin^2 x \cos^2 x\,dx$, write $\sin^2 x \cos^2 x = \tfrac14 \sin^2 2x = \tfrac18(1 - \cos 4x)$, giving $\tfrac{x}{8} - \tfrac{\sin 4x}{32} + C$.

用半角公式计算 $\int \cos^2 x\,dx$。

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

对于 $\int \sin^2 x \cos^2 x\,dx$，写成 $\sin^2 x \cos^2 x = \tfrac14 \sin^2 2x = \tfrac18(1 - \cos 4x)$，得 $\tfrac{x}{8} - \tfrac{\sin 4x}{32} + C$。

Evaluate $\int \tan^3 x \sec^3 x\,dx$. The tangent power is odd, so reserve one factor of $\sec x \tan x$ for the differential and convert $\tan^2 x = \sec^2 x - 1$.

$$\int \tan^2 x \sec^2 x \,(\sec x \tan x)\,dx = \int (\sec^2 x - 1)\sec^2 x\,(\sec x \tan x)\,dx.$$

Let $u = \sec x$, $du = \sec x \tan x\,dx$:

$$\int (u^2-1)u^2\,du = \frac{u^5}{5} - \frac{u^3}{3} + C = \frac{\sec^5 x}{5} - \frac{\sec^3 x}{3} + C.$$

计算 $\int \tan^3 x \sec^3 x\,dx$。正切次数为奇数，故预留一个 $\sec x \tan x$ 因子作为微分，并将 $\tan^2 x = \sec^2 x - 1$ 代入转化。

$$\int \tan^2 x \sec^2 x \,(\sec x \tan x)\,dx = \int (\sec^2 x - 1)\sec^2 x\,(\sec x \tan x)\,dx.$$

令 $u = \sec x$，$du = \sec x \tan x\,dx$：

$$\int (u^2-1)u^2\,du = \frac{u^5}{5} - \frac{u^3}{3} + C = \frac{\sec^5 x}{5} - \frac{\sec^3 x}{3} + C.$$

Evaluate $\int \sin^3 x \cos^3 x\,dx$. Both powers are odd, so either substitution works and the labour is the same. Reserve a factor of $\cos x$ and let $u = \sin x$, $du = \cos x\,dx$, converting $\cos^2 x = 1 - \sin^2 x$.

$$\int \sin^3 x \,(1-\sin^2 x)\,\cos x\,dx = \int u^3(1-u^2)\,du = \int (u^3 - u^5)\,du.$$ $$= \frac{u^4}{4} - \frac{u^6}{6} + C = \frac{\sin^4 x}{4} - \frac{\sin^6 x}{6} + C.$$

Reserving $\sin x$ instead and letting $u=\cos x$ would give $\tfrac{\cos^6 x}{6} - \tfrac{\cos^4 x}{4} + C$. The two answers differ only by a constant, because $\sin^2 x + \cos^2 x = 1$ ties them together; both are correct antiderivatives.

计算 $\int \sin^3 x \cos^3 x\,dx$。两个次数均为奇数，因此两种换元均可行，计算量相同。预留一个 $\cos x$ 因子，令 $u = \sin x$，$du = \cos x\,dx$，将 $\cos^2 x = 1 - \sin^2 x$ 代入。

$$\int \sin^3 x \,(1-\sin^2 x)\,\cos x\,dx = \int u^3(1-u^2)\,du = \int (u^3 - u^5)\,du.$$ $$= \frac{u^4}{4} - \frac{u^6}{6} + C = \frac{\sin^4 x}{4} - \frac{\sin^6 x}{6} + C.$$

若改为预留 $\sin x$ 并令 $u=\cos x$，则得 $\tfrac{\cos^6 x}{6} - \tfrac{\cos^4 x}{4} + C$。两个结果只差一个常数，因为 $\sin^2 x + \cos^2 x = 1$ 将它们联系在一起；两者均是正确的原函数（antiderivative）。

Evaluate $\int \sec^4 x\,dx$. The secant power is even, so reserve one $\sec^2 x$ for the differential of $\tan x$ and convert the remaining $\sec^2 x = 1 + \tan^2 x$.

$$\int \sec^2 x \,\sec^2 x\,dx = \int (1 + \tan^2 x)\,\sec^2 x\,dx.$$

Let $u = \tan x$, $du = \sec^2 x\,dx$:

$$\int (1 + u^2)\,du = u + \frac{u^3}{3} + C = \tan x + \frac{\tan^3 x}{3} + C.$$

The harder relatives $\int \sec x\,dx = \ln|\sec x + \tan x| + C$ and $\int \sec^3 x\,dx = \tfrac12\big(\sec x \tan x + \ln|\sec x + \tan x|\big) + C$ are the base cases reached by the secant reduction formula when no parity trick applies.

计算 $\int \sec^4 x\,dx$。正割次数为偶数，故预留一个 $\sec^2 x$ 作为 $\tan x$ 的微分，并将剩余的 $\sec^2 x = 1 + \tan^2 x$ 代入。

$$\int \sec^2 x \,\sec^2 x\,dx = \int (1 + \tan^2 x)\,\sec^2 x\,dx.$$

令 $u = \tan x$，$du = \sec^2 x\,dx$：

$$\int (1 + u^2)\,du = u + \frac{u^3}{3} + C = \tan x + \frac{\tan^3 x}{3} + C.$$

更难的基本型 $\int \sec x\,dx = \ln|\sec x + \tan x| + C$ 以及 $\int \sec^3 x\,dx = \tfrac12\big(\sec x \tan x + \ln|\sec x + \tan x|\big) + C$，是正割递推公式在没有奇偶技巧可用时达到的基础情形。

Evaluate $\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}}$. Let $x = a\sin\theta$, so $dx = a\cos\theta\,d\theta$ and $\sqrt{a^2 - x^2} = a\cos\theta$ for $-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}$.

$$\int \frac{a\cos\theta\,d\theta}{a\cos\theta} = \int d\theta = \theta + C = \arcsin\!\frac{x}{a} + C.$$

计算 $\displaystyle\int \frac{dx}{\sqrt{a^2 - x^2}}$。令 $x = a\sin\theta$，则 $dx = a\cos\theta\,d\theta$，在 $-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}$ 范围内 $\sqrt{a^2 - x^2} = a\cos\theta$。

$$\int \frac{a\cos\theta\,d\theta}{a\cos\theta} = \int d\theta = \theta + C = \arcsin\!\frac{x}{a} + C.$$

Evaluate $\displaystyle\int \frac{dx}{x^2\sqrt{x^2 + 4}}$. Let $x = 2\tan\theta$, $dx = 2\sec^2\theta\,d\theta$, $\sqrt{x^2+4} = 2\sec\theta$.

$$\int \frac{2\sec^2\theta\,d\theta}{4\tan^2\theta\cdot 2\sec\theta} = \frac14 \int \frac{\sec\theta}{\tan^2\theta}\,d\theta = \frac14 \int \frac{\cos\theta}{\sin^2\theta}\,d\theta.$$

With $w = \sin\theta$ this is $-\tfrac{1}{4\sin\theta} + C$. The reference triangle has opposite $x$, adjacent $2$, hypotenuse $\sqrt{x^2+4}$, so $\sin\theta = x/\sqrt{x^2+4}$ and

$$\int \frac{dx}{x^2\sqrt{x^2+4}} = -\frac{\sqrt{x^2+4}}{4x} + C.$$

计算 $\displaystyle\int \frac{dx}{x^2\sqrt{x^2 + 4}}$。令 $x = 2\tan\theta$，$dx = 2\sec^2\theta\,d\theta$，$\sqrt{x^2+4} = 2\sec\theta$。

$$\int \frac{2\sec^2\theta\,d\theta}{4\tan^2\theta\cdot 2\sec\theta} = \frac14 \int \frac{\sec\theta}{\tan^2\theta}\,d\theta = \frac14 \int \frac{\cos\theta}{\sin^2\theta}\,d\theta.$$

令 $w = \sin\theta$，积分为 $-\tfrac{1}{4\sin\theta} + C$。参考直角三角形对边为 $x$，邻边为 $2$，斜边为 $\sqrt{x^2+4}$，故 $\sin\theta = x/\sqrt{x^2+4}$，从而

$$\int \frac{dx}{x^2\sqrt{x^2+4}} = -\frac{\sqrt{x^2+4}}{4x} + C.$$

Evaluate $\displaystyle\int \sqrt{a^2 - x^2}\,dx$, the integral behind the area of a circle. Let $x = a\sin\theta$, $dx = a\cos\theta\,d\theta$, and $\sqrt{a^2 - x^2} = a\cos\theta$ on $-\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}$ where $\cos\theta\ge 0$.

$$\int a\cos\theta \cdot a\cos\theta\,d\theta = a^2\int \cos^2\theta\,d\theta = a^2\left(\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right) + C.$$

Now $\sin 2\theta = 2\sin\theta\cos\theta = 2\cdot\tfrac{x}{a}\cdot\tfrac{\sqrt{a^2-x^2}}{a}$, and $\theta = \arcsin(x/a)$, so

$$\int \sqrt{a^2 - x^2}\,dx = \frac{a^2}{2}\arcsin\frac{x}{a} + \frac{x\sqrt{a^2 - x^2}}{2} + C.$$

Evaluated from $-a$ to $a$ this gives $\tfrac{a^2}{2}\cdot\pi = \tfrac{\pi a^2}{2}$, the area of the upper half disk of radius $a$, a good sanity check.

计算 $\displaystyle\int \sqrt{a^2 - x^2}\,dx$，这是圆面积背后的积分（integral）。令 $x = a\sin\theta$，$dx = a\cos\theta\,d\theta$，在 $-\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}$ 范围内 $\cos\theta\ge 0$，故 $\sqrt{a^2 - x^2} = a\cos\theta$。

$$\int a\cos\theta \cdot a\cos\theta\,d\theta = a^2\int \cos^2\theta\,d\theta = a^2\left(\frac{\theta}{2} + \frac{\sin 2\theta}{4}\right) + C.$$

注意 $\sin 2\theta = 2\sin\theta\cos\theta = 2\cdot\tfrac{x}{a}\cdot\tfrac{\sqrt{a^2-x^2}}{a}$，且 $\theta = \arcsin(x/a)$，故

$$\int \sqrt{a^2 - x^2}\,dx = \frac{a^2}{2}\arcsin\frac{x}{a} + \frac{x\sqrt{a^2 - x^2}}{2} + C.$$

从 $-a$ 到 $a$ 积分得 $\tfrac{a^2}{2}\cdot\pi = \tfrac{\pi a^2}{2}$，恰好是半径为 $a$ 的上半圆面积，可作为验证。

Evaluate $\displaystyle\int \frac{\sqrt{x^2 - 9}}{x}\,dx$ for $x > 3$. Let $x = 3\sec\theta$, $dx = 3\sec\theta\tan\theta\,d\theta$, and $\sqrt{x^2 - 9} = 3\tan\theta$ on $0 \le \theta < \tfrac{\pi}{2}$.

$$\int \frac{3\tan\theta}{3\sec\theta}\,3\sec\theta\tan\theta\,d\theta = 3\int \tan^2\theta\,d\theta = 3\int (\sec^2\theta - 1)\,d\theta = 3(\tan\theta - \theta) + C.$$

From the triangle, $\tan\theta = \tfrac{\sqrt{x^2-9}}{3}$ and $\theta = \operatorname{arcsec}(x/3) = \arccos(3/x)$, giving

$$\int \frac{\sqrt{x^2-9}}{x}\,dx = \sqrt{x^2 - 9} - 3\arccos\frac{3}{x} + C.$$

计算 $x > 3$ 时的 $\displaystyle\int \frac{\sqrt{x^2 - 9}}{x}\,dx$。令 $x = 3\sec\theta$，$dx = 3\sec\theta\tan\theta\,d\theta$，在 $0 \le \theta < \tfrac{\pi}{2}$ 范围内 $\sqrt{x^2 - 9} = 3\tan\theta$。

$$\int \frac{3\tan\theta}{3\sec\theta}\,3\sec\theta\tan\theta\,d\theta = 3\int \tan^2\theta\,d\theta = 3\int (\sec^2\theta - 1)\,d\theta = 3(\tan\theta - \theta) + C.$$

由三角形可读出 $\tan\theta = \tfrac{\sqrt{x^2-9}}{3}$，$\theta = \operatorname{arcsec}(x/3) = \arccos(3/x)$，故

$$\int \frac{\sqrt{x^2-9}}{x}\,dx = \sqrt{x^2 - 9} - 3\arccos\frac{3}{x} + C.$$

A radical like $\sqrt{x^2 + 2x + 5}$ has no obvious $a^2 \pm u^2$ form until you complete the square: $x^2 + 2x + 5 = (x+1)^2 + 4$. With $u = x+1$ this becomes $\sqrt{u^2 + 4}$, a tangent substitution. In general $ax^2 + bx + c$ is rewritten as $a\big[(x + \tfrac{b}{2a})^2 + \tfrac{4ac - b^2}{4a^2}\big]$, and the sign of $4ac - b^2$ decides which of the three trig substitutions applies. Completing the square is the universal preprocessing step that maps any quadratic radical onto one of the three templates.

To see it through, evaluate $\displaystyle\int \frac{dx}{\sqrt{x^2 + 2x + 5}}$. After $u = x+1$ the integral is $\int \frac{du}{\sqrt{u^2 + 4}}$, a tangent substitution $u = 2\tan\theta$, $du = 2\sec^2\theta\,d\theta$, $\sqrt{u^2+4} = 2\sec\theta$:

$$\int \frac{2\sec^2\theta}{2\sec\theta}\,d\theta = \int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C.$$

Reading the triangle gives $\sec\theta = \tfrac{\sqrt{u^2+4}}{2}$, $\tan\theta = \tfrac{u}{2}$, so the answer is $\ln\big|\sqrt{u^2+4} + u\big| + C'$, that is $\ln\big(\sqrt{x^2+2x+5} + x + 1\big) + C'$, where the additive constant $-\ln 2$ has been absorbed into $C'$. This is exactly the inverse hyperbolic sine $\operatorname{arcsinh}\big(\tfrac{x+1}{2}\big)$ up to a constant, which explains why these integrals are sometimes tabulated with $\operatorname{arcsinh}$ and $\operatorname{arccosh}$ instead of logarithms.

像 $\sqrt{x^2 + 2x + 5}$ 这样的根号，在配方之前看不出 $a^2 \pm u^2$ 的形式：$x^2 + 2x + 5 = (x+1)^2 + 4$。令 $u = x+1$，变为 $\sqrt{u^2 + 4}$，用正切换元。一般地，$ax^2 + bx + c$ 改写为 $a\big[(x + \tfrac{b}{2a})^2 + \tfrac{4ac - b^2}{4a^2}\big]$，$4ac - b^2$ 的符号决定使用三种三角换元中的哪一种。配方是将任意二次根号映射到三种模板之一的通用预处理步骤。

完整演示：计算 $\displaystyle\int \frac{dx}{\sqrt{x^2 + 2x + 5}}$。令 $u = x+1$，积分（integral）变为 $\int \frac{du}{\sqrt{u^2 + 4}}$，用正切换元 $u = 2\tan\theta$，$du = 2\sec^2\theta\,d\theta$，$\sqrt{u^2+4} = 2\sec\theta$：

$$\int \frac{2\sec^2\theta}{2\sec\theta}\,d\theta = \int \sec\theta\,d\theta = \ln|\sec\theta + \tan\theta| + C.$$

由三角形读出 $\sec\theta = \tfrac{\sqrt{u^2+4}}{2}$，$\tan\theta = \tfrac{u}{2}$，故结果为 $\ln\big|\sqrt{u^2+4} + u\big| + C'$，即 $\ln\big(\sqrt{x^2+2x+5} + x + 1\big) + C'$，其中加法常数 $-\ln 2$ 已并入 $C'$。这等价于（相差一个常数）反双曲正弦 $\operatorname{arcsinh}\big(\tfrac{x+1}{2}\big)$，解释了为何某些积分表用 $\operatorname{arcsinh}$ 和 $\operatorname{arccosh}$ 代替对数来列表。

Evaluate $\displaystyle\int \frac{5x - 4}{x^2 - x - 2}\,dx$. Factor the denominator: $x^2 - x - 2 = (x-2)(x+1)$. Set

$$\frac{5x-4}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}, \qquad 5x - 4 = A(x+1) + B(x-2).$$

At $x = 2$: $6 = 3A$, so $A = 2$. At $x = -1$: $-9 = -3B$, so $B = 3$. Therefore

$$\int \left(\frac{2}{x-2} + \frac{3}{x+1}\right)dx = 2\ln|x-2| + 3\ln|x+1| + C.$$

计算 $\displaystyle\int \frac{5x - 4}{x^2 - x - 2}\,dx$。分解分母：$x^2 - x - 2 = (x-2)(x+1)$，设

$$\frac{5x-4}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}, \qquad 5x - 4 = A(x+1) + B(x-2).$$

令 $x = 2$：$6 = 3A$，故 $A = 2$。令 $x = -1$：$-9 = -3B$，故 $B = 3$。因此

$$\int \left(\frac{2}{x-2} + \frac{3}{x+1}\right)dx = 2\ln|x-2| + 3\ln|x+1| + C.$$

Evaluate $\displaystyle\int \frac{x^2 + 1}{x^2 - 1}\,dx$. The fraction is improper, so divide: $\frac{x^2+1}{x^2-1} = 1 + \frac{2}{x^2-1}$. Now decompose the proper remainder over $(x-1)(x+1)$:

$$\frac{2}{(x-1)(x+1)} = \frac{1}{x-1} - \frac{1}{x+1}.$$ $$\int \left(1 + \frac{1}{x-1} - \frac{1}{x+1}\right)dx = x + \ln|x-1| - \ln|x+1| + C.$$

计算 $\displaystyle\int \frac{x^2 + 1}{x^2 - 1}\,dx$。分式是假分式，先做除法：$\frac{x^2+1}{x^2-1} = 1 + \frac{2}{x^2-1}$。再将真分式余项在 $(x-1)(x+1)$ 上展开：

$$\frac{2}{(x-1)(x+1)} = \frac{1}{x-1} - \frac{1}{x+1}.$$ $$\int \left(1 + \frac{1}{x-1} - \frac{1}{x+1}\right)dx = x + \ln|x-1| - \ln|x+1| + C.$$

Evaluate $\displaystyle\int \frac{x^2 + 1}{x(x-1)(x+2)}\,dx$. The fraction is proper (degree 2 over degree 3), so set

$$\frac{x^2+1}{x(x-1)(x+2)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+2}.$$

Cover-up at each root. At $x=0$: $A = \dfrac{0+1}{(0-1)(0+2)} = \dfrac{1}{-2} = -\tfrac12$. At $x=1$: $B = \dfrac{1+1}{(1)(1+2)} = \dfrac{2}{3}$. At $x=-2$: $C = \dfrac{4+1}{(-2)(-2-1)} = \dfrac{5}{6}$.

$$\int \left(\frac{-1/2}{x} + \frac{2/3}{x-1} + \frac{5/6}{x+2}\right)dx = -\frac12\ln|x| + \frac23\ln|x-1| + \frac56\ln|x+2| + C.$$

计算 $\displaystyle\int \frac{x^2 + 1}{x(x-1)(x+2)}\,dx$。分式为真分式（分子次数 2 低于分母次数 3），设

$$\frac{x^2+1}{x(x-1)(x+2)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+2}.$$

在每个根处应用覆盖法。$x=0$：$A = \dfrac{0+1}{(0-1)(0+2)} = \dfrac{1}{-2} = -\tfrac12$。$x=1$：$B = \dfrac{1+1}{(1)(1+2)} = \dfrac{2}{3}$。$x=-2$：$C = \dfrac{4+1}{(-2)(-2-1)} = \dfrac{5}{6}$。

$$\int \left(\frac{-1/2}{x} + \frac{2/3}{x-1} + \frac{5/6}{x+2}\right)dx = -\frac12\ln|x| + \frac23\ln|x-1| + \frac56\ln|x+2| + C.$$

Evaluate $\displaystyle\int_3^4 \frac{dx}{x^2 - 4}$. Factor $x^2 - 4 = (x-2)(x+2)$ and decompose $\dfrac{1}{(x-2)(x+2)} = \dfrac{1/4}{x-2} - \dfrac{1/4}{x+2}$.

$$\int_3^4 \frac{dx}{x^2-4} = \frac14\Big[\ln|x-2| - \ln|x+2|\Big]_3^4 = \frac14\ln\frac{x-2}{x+2}\bigg|_3^4.$$ $$= \frac14\left(\ln\frac{2}{6} - \ln\frac{1}{5}\right) = \frac14\ln\frac{2/6}{1/5} = \frac14\ln\frac{5}{3}.$$

Because the interval $[3,4]$ avoids the poles at $\pm 2$, the integrand stays continuous and the answer is a genuine number, here $\tfrac14\ln\tfrac53 \approx 0.1277$.

计算 $\displaystyle\int_3^4 \frac{dx}{x^2 - 4}$。分解 $x^2 - 4 = (x-2)(x+2)$，再展开 $\dfrac{1}{(x-2)(x+2)} = \dfrac{1/4}{x-2} - \dfrac{1/4}{x+2}$。

$$\int_3^4 \frac{dx}{x^2-4} = \frac14\Big[\ln|x-2| - \ln|x+2|\Big]_3^4 = \frac14\ln\frac{x-2}{x+2}\bigg|_3^4.$$ $$= \frac14\left(\ln\frac{2}{6} - \ln\frac{1}{5}\right) = \frac14\ln\frac{2/6}{1/5} = \frac14\ln\frac{5}{3}.$$

由于区间 $[3,4]$ 远离极点 $\pm 2$，被积函数保持连续，答案为具体数值，此处 $\tfrac14\ln\tfrac53 \approx 0.1277$。

Decompose $\displaystyle\frac{x}{(x+1)^2}$. Write $\dfrac{x}{(x+1)^2} = \dfrac{A}{x+1} + \dfrac{B}{(x+1)^2}$, so $x = A(x+1) + B$. At $x = -1$: $-1 = B$. Matching the $x$ coefficient: $A = 1$. Then

$$\int \frac{x}{(x+1)^2}\,dx = \int \left(\frac{1}{x+1} - \frac{1}{(x+1)^2}\right)dx = \ln|x+1| + \frac{1}{x+1} + C.$$

分解 $\displaystyle\frac{x}{(x+1)^2}$。设 $\dfrac{x}{(x+1)^2} = \dfrac{A}{x+1} + \dfrac{B}{(x+1)^2}$，则 $x = A(x+1) + B$。令 $x = -1$：$-1 = B$。比较 $x$ 的系数：$A = 1$。于是

$$\int \frac{x}{(x+1)^2}\,dx = \int \left(\frac{1}{x+1} - \frac{1}{(x+1)^2}\right)dx = \ln|x+1| + \frac{1}{x+1} + C.$$

Evaluate $\displaystyle\int \frac{2x + 3}{x^2 + 4}\,dx$. The denominator is irreducible. Split the numerator into a piece proportional to the derivative $2x$ of the denominator and a constant:

$$\int \frac{2x}{x^2+4}\,dx + \int \frac{3}{x^2+4}\,dx = \ln(x^2+4) + \frac{3}{2}\arctan\frac{x}{2} + C.$$

计算 $\displaystyle\int \frac{2x + 3}{x^2 + 4}\,dx$。分母不可约，将分子拆分为与分母导数 $2x$ 成比例的部分和一个常数：

$$\int \frac{2x}{x^2+4}\,dx + \int \frac{3}{x^2+4}\,dx = \ln(x^2+4) + \frac{3}{2}\arctan\frac{x}{2} + C.$$

Evaluate $\displaystyle\int \frac{x^2 + 2x + 4}{(x-1)(x^2+1)}\,dx$. The denominator has one linear factor and one irreducible quadratic, so

$$\frac{x^2+2x+4}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}, \quad x^2+2x+4 = A(x^2+1) + (Bx+C)(x-1).$$

At $x=1$: $7 = 2A$, so $A = \tfrac72$. Match $x^2$ coefficients: $1 = A + B$, so $B = 1 - \tfrac72 = -\tfrac52$. Match constants: $4 = A - C$, so $C = A - 4 = -\tfrac12$. Thus

$$\int \left(\frac{7/2}{x-1} + \frac{-\tfrac52 x - \tfrac12}{x^2+1}\right)dx = \frac72\ln|x-1| - \frac54\ln(x^2+1) - \frac12\arctan x + C.$$

The quadratic term split as $-\tfrac52\int\tfrac{x}{x^2+1}\,dx - \tfrac12\int\tfrac{dx}{x^2+1}$, the first piece giving a logarithm and the second an arctangent, exactly the two outputs an irreducible quadratic always produces.

计算 $\displaystyle\int \frac{x^2 + 2x + 4}{(x-1)(x^2+1)}\,dx$。分母含一个线性因子和一个不可约二次因子，设

$$\frac{x^2+2x+4}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}, \quad x^2+2x+4 = A(x^2+1) + (Bx+C)(x-1).$$

令 $x=1$：$7 = 2A$，故 $A = \tfrac72$。比较 $x^2$ 系数：$1 = A + B$，故 $B = -\tfrac52$。比较常数项：$4 = A - C$，故 $C = -\tfrac12$。因此

$$\int \left(\frac{7/2}{x-1} + \frac{-\tfrac52 x - \tfrac12}{x^2+1}\right)dx = \frac72\ln|x-1| - \frac54\ln(x^2+1) - \frac12\arctan x + C.$$

二次项拆为 $-\tfrac52\int\tfrac{x}{x^2+1}\,dx - \tfrac12\int\tfrac{dx}{x^2+1}$，第一项给出对数，第二项给出反正切，这正是不可约二次因子必然产生的两类输出。

Evaluate $\displaystyle\int \frac{dx}{x^2 + 2x + 5}$. The denominator is irreducible (discriminant $4 - 20 < 0$). Complete the square: $x^2 + 2x + 5 = (x+1)^2 + 4$. With $u = x+1$,

$$\int \frac{du}{u^2 + 4} = \frac12\arctan\frac{u}{2} + C = \frac12\arctan\frac{x+1}{2} + C.$$

If the numerator had been linear, say $\int\tfrac{x}{x^2+2x+5}\,dx$, you would first write $x = \tfrac12(2x+2) - 1$ to align part of the numerator with the derivative $2x+2$ of the denominator, producing a logarithm plus this arctangent.

计算 $\displaystyle\int \frac{dx}{x^2 + 2x + 5}$。分母不可约（判别式 $4 - 20 < 0$），配方得 $x^2 + 2x + 5 = (x+1)^2 + 4$。令 $u = x+1$：

$$\int \frac{du}{u^2 + 4} = \frac12\arctan\frac{u}{2} + C = \frac12\arctan\frac{x+1}{2} + C.$$

若分子为线性，例如 $\int\tfrac{x}{x^2+2x+5}\,dx$，则先将 $x$ 写成 $\tfrac12(2x+2) - 1$，使分子一部分与分母的导数 $2x+2$ 对齐，产生一个对数项加上此反正切。

By the fundamental theorem of algebra, any real polynomial factors over the reals into linear factors $(x-r)$ and irreducible quadratics $x^2 + px + q$. The partial fraction theorem then guarantees that a proper rational function is a finite sum of terms of exactly four shapes:

$$\frac{A}{(x-r)^k}, \qquad \frac{Bx + C}{(x^2+px+q)^k}.$$

Each shape integrates in closed form. Powers of linear factors give powers and logarithms: $\int (x-r)^{-1}\,dx = \ln|x-r|$ and $\int (x-r)^{-k}\,dx = \tfrac{(x-r)^{1-k}}{1-k}$ for $k>1$. For the quadratic term, complete the square so that $x^2 + px + q = (x + \tfrac{p}{2})^2 + a^2$ with $a^2 = q - \tfrac{p^2}{4} > 0$, and split the numerator into a multiple of the derivative $2x + p$ plus a constant:

$$\int \frac{Bx + C}{x^2 + px + q}\,dx = \frac{B}{2}\ln(x^2+px+q) + \left(C - \frac{Bp}{2}\right)\cdot\frac{1}{a}\arctan\frac{x + p/2}{a} + \text{const}.$$

The first piece is a logarithm because the numerator was engineered to be the denominator's derivative; the leftover constant numerator over $(x+\tfrac{p}{2})^2 + a^2$ is precisely the arctangent template. Higher powers $\int \tfrac{dx}{(u^2+a^2)^k}$ are handled by the reduction formula

$$\int \frac{du}{(u^2+a^2)^k} = \frac{u}{2a^2(k-1)(u^2+a^2)^{k-1}} + \frac{2k-3}{2a^2(k-1)}\int \frac{du}{(u^2+a^2)^{k-1}},$$

which steps every multiplicity down to the base arctangent. Hence every proper rational function has an elementary antiderivative, a result with no analogue for general elementary integrands.

由代数基本定理，任何实系数多项式在实数范围内均可分解为线性因子 $(x-r)$ 和不可约二次式 $x^2 + px + q$ 的乘积。部分分式定理随即保证：真分式是有限个以下四种形式之和：

$$\frac{A}{(x-r)^k}, \qquad \frac{Bx + C}{(x^2+px+q)^k}.$$

每种形式都有初等原函数（antiderivative）。线性因子的幂次给出幂函数和对数：$\int (x-r)^{-1}\,dx = \ln|x-r|$，$k>1$ 时 $\int (x-r)^{-k}\,dx = \tfrac{(x-r)^{1-k}}{1-k}$。对二次项，配方使 $x^2 + px + q = (x + \tfrac{p}{2})^2 + a^2$（其中 $a^2 = q - \tfrac{p^2}{4} > 0$），再将分子拆为分母导数 $2x + p$ 的倍数加一个常数：

$$\int \frac{Bx + C}{x^2 + px + q}\,dx = \frac{B}{2}\ln(x^2+px+q) + \left(C - \frac{Bp}{2}\right)\cdot\frac{1}{a}\arctan\frac{x + p/2}{a} + \text{const}.$$

第一项为对数，因为分子被设计成与分母的导数成比例；剩下的常数分子除以 $(x+\tfrac{p}{2})^2 + a^2$ 恰好是反正切模板。更高幂次 $\int \tfrac{dx}{(u^2+a^2)^k}$ 由递推公式处理：

$$\int \frac{du}{(u^2+a^2)^k} = \frac{u}{2a^2(k-1)(u^2+a^2)^{k-1}} + \frac{2k-3}{2a^2(k-1)}\int \frac{du}{(u^2+a^2)^{k-1}},$$

它将每个幂次逐步降至基础反正切情形。因此每个真分式都有初等原函数，这一结论对一般初等被积函数并不成立。

Evaluate $\displaystyle\int \frac{dx}{1 + \sqrt{x}}$. Let $u = \sqrt{x}$, so $x = u^2$ and $dx = 2u\,du$.

$$\int \frac{2u}{1+u}\,du = 2\int \left(1 - \frac{1}{1+u}\right)du = 2u - 2\ln|1+u| + C.$$ $$= 2\sqrt{x} - 2\ln(1 + \sqrt{x}) + C.$$

计算 $\displaystyle\int \frac{dx}{1 + \sqrt{x}}$。令 $u = \sqrt{x}$，则 $x = u^2$，$dx = 2u\,du$。

$$\int \frac{2u}{1+u}\,du = 2\int \left(1 - \frac{1}{1+u}\right)du = 2u - 2\ln|1+u| + C.$$ $$= 2\sqrt{x} - 2\ln(1 + \sqrt{x}) + C.$$

Evaluate $\displaystyle\int \frac{dx}{1 + \sin x}$. With $t = \tan(x/2)$ the integrand rationalizes:

$$\int \frac{\frac{2}{1+t^2}\,dt}{1 + \frac{2t}{1+t^2}} = \int \frac{2\,dt}{1 + t^2 + 2t} = \int \frac{2\,dt}{(1+t)^2} = -\frac{2}{1+t} + C.$$ $$= -\frac{2}{1 + \tan(x/2)} + C.$$

计算 $\displaystyle\int \frac{dx}{1 + \sin x}$。令 $t = \tan(x/2)$，被积函数有理化：

$$\int \frac{\frac{2}{1+t^2}\,dt}{1 + \frac{2t}{1+t^2}} = \int \frac{2\,dt}{1 + t^2 + 2t} = \int \frac{2\,dt}{(1+t)^2} = -\frac{2}{1+t} + C.$$ $$= -\frac{2}{1 + \tan(x/2)} + C.$$

Evaluate $\displaystyle\int \frac{dx}{\sqrt{x} + \sqrt[3]{x}}$. The two roots share the least common index $6$, so set $u = x^{1/6}$, giving $x = u^6$, $dx = 6u^5\,du$, $\sqrt{x} = u^3$, $\sqrt[3]{x} = u^2$.

$$\int \frac{6u^5}{u^3 + u^2}\,du = 6\int \frac{u^5}{u^2(u+1)}\,du = 6\int \frac{u^3}{u+1}\,du.$$

Polynomial-divide $\dfrac{u^3}{u+1} = u^2 - u + 1 - \dfrac{1}{u+1}$:

$$6\int \left(u^2 - u + 1 - \frac{1}{u+1}\right)du = 6\left(\frac{u^3}{3} - \frac{u^2}{2} + u - \ln|u+1|\right) + C.$$ $$= 2\sqrt{x} - 3\sqrt[3]{x} + 6\sqrt[6]{x} - 6\ln\big(\sqrt[6]{x} + 1\big) + C.$$

计算 $\displaystyle\int \frac{dx}{\sqrt{x} + \sqrt[3]{x}}$。两个根号的最小公共指数为 $6$，故令 $u = x^{1/6}$，得 $x = u^6$，$dx = 6u^5\,du$，$\sqrt{x} = u^3$，$\sqrt[3]{x} = u^2$。

$$\int \frac{6u^5}{u^3 + u^2}\,du = 6\int \frac{u^5}{u^2(u+1)}\,du = 6\int \frac{u^3}{u+1}\,du.$$

多项式除法：$\dfrac{u^3}{u+1} = u^2 - u + 1 - \dfrac{1}{u+1}$：

$$6\int \left(u^2 - u + 1 - \frac{1}{u+1}\right)du = 6\left(\frac{u^3}{3} - \frac{u^2}{2} + u - \ln|u+1|\right) + C.$$ $$= 2\sqrt{x} - 3\sqrt[3]{x} + 6\sqrt[6]{x} - 6\ln\big(\sqrt[6]{x} + 1\big) + C.$$

Evaluate $\displaystyle\int \frac{dx}{2 + \cos x}$. With $t = \tan(x/2)$, $\cos x = \tfrac{1-t^2}{1+t^2}$ and $dx = \tfrac{2}{1+t^2}\,dt$:

$$\int \frac{\frac{2}{1+t^2}}{2 + \frac{1-t^2}{1+t^2}}\,dt = \int \frac{2\,dt}{2(1+t^2) + (1-t^2)} = \int \frac{2\,dt}{t^2 + 3}.$$ $$= \frac{2}{\sqrt3}\arctan\frac{t}{\sqrt3} + C = \frac{2}{\sqrt3}\arctan\!\left(\frac{\tan(x/2)}{\sqrt3}\right) + C.$$

The Weierstrass substitution converts any rational function of $\sin x$ and $\cos x$ into a rational function of $t$, which then falls to partial fractions, so it is the universal fallback for trigonometric rational integrands.

计算 $\displaystyle\int \frac{dx}{2 + \cos x}$。令 $t = \tan(x/2)$，$\cos x = \tfrac{1-t^2}{1+t^2}$，$dx = \tfrac{2}{1+t^2}\,dt$：

$$\int \frac{\frac{2}{1+t^2}}{2 + \frac{1-t^2}{1+t^2}}\,dt = \int \frac{2\,dt}{2(1+t^2) + (1-t^2)} = \int \frac{2\,dt}{t^2 + 3}.$$ $$= \frac{2}{\sqrt3}\arctan\frac{t}{\sqrt3} + C = \frac{2}{\sqrt3}\arctan\!\left(\frac{\tan(x/2)}{\sqrt3}\right) + C.$$

Weierstrass 换元可将任何关于 $\sin x$ 和 $\cos x$ 的有理函数化为 $t$ 的有理函数，再利用部分分式（partial fractions）完成积分，因此它是三角有理被积函数的万能后备方案。

Consider $\displaystyle\int \frac{x}{\sqrt{1 - x^2}}\,dx$. There is a quadratic radical, which would suggest trig substitution, but first check for a plain $u$-substitution: the numerator $x$ is, up to a constant, the derivative of $1 - x^2$ inside the radical. Let $u = 1 - x^2$, $du = -2x\,dx$:

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

The lesson: always test the cheaper substitution before committing to trig substitution.

考虑 $\displaystyle\int \frac{x}{\sqrt{1 - x^2}}\,dx$。含二次根号，本能地想到三角换元，但先验证是否有简单的 $u$ 换元：分子 $x$ 恰好（相差一个常数）是根号内 $1 - x^2$ 的导数。令 $u = 1 - x^2$，$du = -2x\,dx$：

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

结论：在决定使用三角换元之前，务必先检验更简单的换元是否可行。

Evaluate $\displaystyle\int \frac{x^3}{\sqrt{x^2 + 1}}\,dx$. The integrand mixes a polynomial and a radical. Substitute $u = x^2 + 1$, so $x^2 = u - 1$ and $x\,dx = \tfrac12\,du$:

$$\int \frac{x^2 \cdot x}{\sqrt{x^2+1}}\,dx = \frac12\int \frac{u-1}{\sqrt{u}}\,du = \frac12\int \left(u^{1/2} - u^{-1/2}\right)du.$$ $$= \frac12\left(\frac{2}{3}u^{3/2} - 2u^{1/2}\right) + C = \frac13(x^2+1)^{3/2} - (x^2+1)^{1/2} + C.$$

计算 $\displaystyle\int \frac{x^3}{\sqrt{x^2 + 1}}\,dx$。被积函数混合了多项式和根号。令 $u = x^2 + 1$，则 $x^2 = u - 1$，$x\,dx = \tfrac12\,du$：

$$\int \frac{x^2 \cdot x}{\sqrt{x^2+1}}\,dx = \frac12\int \frac{u-1}{\sqrt{u}}\,du = \frac12\int \left(u^{1/2} - u^{-1/2}\right)du.$$ $$= \frac12\left(\frac{2}{3}u^{3/2} - 2u^{1/2}\right) + C = \frac13(x^2+1)^{3/2} - (x^2+1)^{1/2} + C.$$

Evaluate $\displaystyle\int \frac{dx}{1 + e^x}$. No standard pattern fits until you rewrite. Multiply top and bottom by $e^{-x}$:

$$\int \frac{e^{-x}}{e^{-x} + 1}\,dx.$$

Now the numerator is, up to sign, the derivative of the denominator. Let $u = e^{-x} + 1$, $du = -e^{-x}\,dx$:

$$-\int \frac{du}{u} = -\ln|u| + C = -\ln(e^{-x} + 1) + C = x - \ln(e^x + 1) + C,$$

using $-\ln(e^{-x}+1) = -\ln\big(e^{-x}(1 + e^x)\big) = x - \ln(1+e^x)$. The lesson of the strategy section is exactly this: many integrals only reveal a usable substitution after an algebraic rewrite.

计算 $\displaystyle\int \frac{dx}{1 + e^x}$。改写前没有标准模式可用。分子分母同乘 $e^{-x}$：

$$\int \frac{e^{-x}}{e^{-x} + 1}\,dx.$$

此时分子（相差一个符号）恰好是分母的导数。令 $u = e^{-x} + 1$，$du = -e^{-x}\,dx$：

$$-\int \frac{du}{u} = -\ln|u| + C = -\ln(e^{-x} + 1) + C = x - \ln(e^x + 1) + C,$$

利用 $-\ln(e^{-x}+1) = -\ln\big(e^{-x}(1 + e^x)\big) = x - \ln(1+e^x)$。这正是积分策略的核心：许多积分只有在代数改写后才显现出可用的换元。

Evaluate $\displaystyle\int x\arctan x\,dx$. A product of a polynomial with an inverse trig function points to integration by parts, with $u = \arctan x$, $dv = x\,dx$, so $du = \tfrac{dx}{1+x^2}$, $v = \tfrac{x^2}{2}$:

$$\frac{x^2}{2}\arctan x - \frac12\int \frac{x^2}{1 + x^2}\,dx = \frac{x^2}{2}\arctan x - \frac12\int \left(1 - \frac{1}{1+x^2}\right)dx.$$ $$= \frac{x^2}{2}\arctan x - \frac{x}{2} + \frac12\arctan x + C.$$

The leftover $\int \tfrac{x^2}{1+x^2}\,dx$ was improper as a rational function, so the strategy looped back to the divide-first rule of Section 3 before finishing with an arctangent.

计算 $\displaystyle\int x\arctan x\,dx$。多项式与反三角函数的乘积提示分部积分，令 $u = \arctan x$，$dv = x\,dx$，则 $du = \tfrac{dx}{1+x^2}$，$v = \tfrac{x^2}{2}$：

$$\frac{x^2}{2}\arctan x - \frac12\int \frac{x^2}{1 + x^2}\,dx = \frac{x^2}{2}\arctan x - \frac12\int \left(1 - \frac{1}{1+x^2}\right)dx.$$ $$= \frac{x^2}{2}\arctan x - \frac{x}{2} + \frac12\arctan x + C.$$

剩余的 $\int \tfrac{x^2}{1+x^2}\,dx$ 是假分式，策略返回第 3 节的"先除法"规则，最终以反正切收尾。

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

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

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

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

The term $\int \cos^n x\,dx$ reappears on the right. Collecting it on the left gives $n\int \cos^n x\,dx = \cos^{n-1}x \sin x + (n-1)\int \cos^{n-2}x\,dx$, which is the stated formula after dividing by $n$.

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

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

代入 $\sin^2 x = 1 - \cos^2 x$：

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

右边重新出现 $\int \cos^n x\,dx$，将其移至左边，得 $n\int \cos^n x\,dx = \cos^{n-1}x \sin x + (n-1)\int \cos^{n-2}x\,dx$，除以 $n$ 即得所述公式。

The tangent reduction needs no integration by parts, only the Pythagorean identity. Split off two factors and substitute $\tan^2 x = \sec^2 x - 1$:

$$\int \tan^n x\,dx = \int \tan^{n-2}x\,(\sec^2 x - 1)\,dx = \int \tan^{n-2}x\,\sec^2 x\,dx - \int \tan^{n-2}x\,dx.$$

In the first integral let $u = \tan x$, $du = \sec^2 x\,dx$, so it equals $\tfrac{\tan^{n-1}x}{n-1}$. Therefore

$$\int \tan^n x\,dx = \frac{\tan^{n-1}x}{n-1} - \int \tan^{n-2}x\,dx,$$

which steps the power down by two each time until it reaches $\int \tan x\,dx = \ln|\sec x| + C$ (odd $n$) or $\int dx = x + C$ (even $n$). For example, $\int \tan^3 x\,dx = \tfrac{\tan^2 x}{2} - \int \tan x\,dx = \tfrac{\tan^2 x}{2} - \ln|\sec x| + C$. Every reduction formula in this unit works the same way: an identity exposes a self-similar copy of the original integral, which is then solved algebraically.

正切递推公式无需分部积分，只需勾股恒等式。分离两个因子并代入 $\tan^2 x = \sec^2 x - 1$：

$$\int \tan^n x\,dx = \int \tan^{n-2}x\,(\sec^2 x - 1)\,dx = \int \tan^{n-2}x\,\sec^2 x\,dx - \int \tan^{n-2}x\,dx.$$

第一个积分令 $u = \tan x$，$du = \sec^2 x\,dx$，得 $\tfrac{\tan^{n-1}x}{n-1}$。因此

$$\int \tan^n x\,dx = \frac{\tan^{n-1}x}{n-1} - \int \tan^{n-2}x\,dx,$$

每次将次数降低 2，直到 $\int \tan x\,dx = \ln|\sec x| + C$（$n$ 为奇数）或 $\int dx = x + C$（$n$ 为偶数）。例如 $\int \tan^3 x\,dx = \tfrac{\tan^2 x}{2} - \int \tan x\,dx = \tfrac{\tan^2 x}{2} - \ln|\sec x| + C$。本单元所有递推公式的原理相同：某个恒等式将原积分的一个自相似副本暴露出来，再代数求解即得。

Every technique here produces an elementary antiderivative, a function built from polynomials, roots, exponentials, logarithms, and trig functions and their inverses. It is tempting to assume every continuous function has such an antiderivative, but this is false. Liouville's theorem, made effective by the Risch algorithm, shows that

$$\int e^{-x^2}\,dx, \qquad \int \frac{\sin x}{x}\,dx, \qquad \int \frac{dx}{\ln x}$$

have no elementary closed form. These integrals are perfectly well defined (they name genuine functions, such as the error function and the logarithmic integral), but no finite combination of elementary operations expresses them. The methods of this unit therefore solve a large but proper subset of integration problems, and recognizing when an integral falls outside that subset is itself a mark of fluency.

The distinction is between an indefinite and a definite integral. Although $\int e^{-x^2}\,dx$ has no elementary antiderivative, the definite integral over the whole line is exactly computable by a multivariable trick: squaring $I = \int_{-\infty}^{\infty} e^{-x^2}\,dx$ and switching to polar coordinates gives $I^2 = \int_0^{2\pi}\int_0^\infty e^{-r^2}r\,dr\,d\theta = \pi$, so $I = \sqrt\pi$. The reduction and parity methods of this unit also power the Wallis formulas, the closed forms for $\int_0^{\pi/2}\sin^n x\,dx$, which are the workhorse of probability and Fourier analysis. So even where a single antiderivative is out of reach, the techniques here remain the right starting point for the definite integrals that occur in practice.

本处每种技巧都给出初等原函数，即由多项式、根号、指数函数、对数以及三角函数及其反函数构成的函数。人们容易以为每个连续函数都有这样的原函数，但这是错误的。刘维尔定理（由 Risch 算法实现有效化）表明

$$\int e^{-x^2}\,dx, \qquad \int \frac{\sin x}{x}\,dx, \qquad \int \frac{dx}{\ln x}$$

均没有初等封闭形式。这些积分（integral）定义完全良好（它们命名了真实的函数，如误差函数和对数积分），但无法用有限个初等运算表达。本单元的方法因此只解决了一个大而真实的积分子集，而识别一个积分何时超出该子集本身就是数学流畅度的标志。

不定积分与定积分之间存在重要区别。虽然 $\int e^{-x^2}\,dx$ 没有初等原函数，但全直线上的定积分可用多变量技巧精确计算：对 $I = \int_{-\infty}^{\infty} e^{-x^2}\,dx$ 取平方并转换为极坐标，得 $I^2 = \int_0^{2\pi}\int_0^\infty e^{-r^2}r\,dr\,d\theta = \pi$，故 $I = \sqrt\pi$。本单元的递推和奇偶方法还支撑了 Wallis 公式，即 $\int_0^{\pi/2}\sin^n x\,dx$ 的封闭形式，这是概率论和傅里叶分析的基本工具。因此即便在单个原函数无法企及之处，本单元的技巧仍是处理实践中出现的定积分的正确起点。