习题课讲解

8.2 习题课讲解

8.2.1 有理函数的不定积分

例 8.2.1 计算以下不定积分：

(1): $\int \frac{1 - x^{2}}{1 + x^{2}} d x$
(2): $\int \frac{x}{3 - x^{2}} d x$
(3): $\int \frac{2 x + 1}{x^{2} + x + 1} d x$
(4): $\int \frac{x}{x^{2} + x - 6} d x$
(5): $\int \frac{x^{2}}{1 + x^{6}} d x$
(6): $\int \frac{x - 1}{x^{2} - 4 x + 8} d x$
(7): $\int \frac{1}{(x + 1) (x + 2)} d x$
(8): $\int \frac{1}{x (1 + x^{2})} d x$
(9): $\int \frac{x^{3} + 1}{x^{3} - 5 x^{2} + 6 x} d x$
(10): $\int \frac{1}{x^{4} - 1} d x$
(11): $\int \frac{x^{4}}{x^{4} + 5 x^{2} + 4} d x$
(12): $\int \frac{1}{1 + x^{3}} d x$
(13): $\int \frac{x^{7}}{{(1 - x^{2})}^{5}} d x$
(14): $\int \frac{1}{x^{4} (2 x^{2} - 1)} d x$
(15): $\int \frac{1}{x (x^{n} + a)} d x$

解 (1) 线性、观察原函数 $\begin{matrix} (1) & \int \frac{1 - x^{2}}{1 + x^{2}} d x = \int \frac{2}{1 + x^{2}} d x - \int d x = 2 \arctan x - x + C \end{matrix}$

(2) 凑微分 $\begin{matrix} (2) & \int \frac{x}{3 - x^{2}} d x = \frac{1}{2} \int \frac{1}{3 - x^{2}} d (x^{2} - 3) = - \frac{1}{2} \ln | x^{2} - 3 | + C \end{matrix}$

(3) 凑微分 $\begin{matrix} (3) & \int \frac{2 x + 1}{x^{2} + x + 1} d x = \int \frac{1}{x^{2} + x + 1} d (x^{2} + x + 1) = \ln | x^{2} + x + 1 | + C \end{matrix}$

(4) 最简因式分解 $\begin{matrix} (4) & \int \frac{x}{x^{2} + x - 6} d x = \frac{1}{5} \int (\frac{3}{x - 3} + \frac{2}{x + 2}) d x = \frac{3}{5} \ln | x - 3 | + \frac{2}{5} \ln | x + 2 | + C \end{matrix}$

(5) 凑微分 $\begin{matrix} (5) & \int \frac{x^{2}}{1 + x^{6}} d x = \frac{1}{3} \int \frac{1}{{(x^{3})}^{2} + 1} d x^{3} = \frac{1}{3} \arctan x^{3} + C \end{matrix}$

(6) 线性、换元 $\begin{matrix} (6) & \begin{aligned} \int \frac{x - 1}{x^{2} - 4 x + 8} d x & = \int \frac{x - 2}{(x - 2)^{2} + 4} d x + \int \frac{1}{(x - 2)^{2} + 4} d x \\ = \frac{1}{2} \int \frac{1}{u} d u + \frac{1}{2} \int \frac{1}{v^{2} + 1} d v, u = (x - 2)^{2} + 4, v = \frac{x - 2}{2} \\ = \frac{1}{2} \ln | x^{2} - 4 x + 8 | + \arctan \frac{x - 2}{2} + C \end{aligned} \end{matrix}$

(7) 最简分式分解 $\begin{matrix} (7) & \int \frac{1}{(x + 1) (x + 2)} d x = \int (\frac{1}{x + 1} - \frac{1}{x + 2}) d x = \ln | \frac{x + 1}{x + 2} | + C \end{matrix}$

(8) 凑微分、线性 $\begin{matrix} (8) & \int \frac{1}{x (1 + x^{2})} d x = \frac{1}{2} \int \frac{1}{x^{2} (1 + x^{2})} d x^{2} = \ln | x | - \frac{1}{2} \ln (1 + x^{2}) + C \end{matrix}$

(9) 设 $\begin{matrix} (9) & 5 x^{2} - 6 x + 1 = A (x - 2) (x - 3) + B x (x - 3) + E x (x - 2) \end{matrix}$ 则 $\begin{matrix} (10) & \int \frac{x^{3} + 1}{x^{3} - 5 x^{2} + 6 x} d x = \int [1 + \frac{5 x^{2} - 6 x + 1}{x (x - 2) (x - 3)}] d x = x + \int (\frac{A}{x} + \frac{B}{x - 2} + \frac{E}{x - 3}) d x \end{matrix}$ 两边取 $x = 0$ ，得到 $A = \frac{1}{6}$ ；取 $x = 2$ ，得到 $B = - \frac{9}{2}$ ；取 $x = 3$ ，得到 $E = \frac{28}{3}$ 。故 $\begin{matrix} (11) & \int \frac{x^{3} + 1}{x^{3} - 5 x^{2} + 6 x} d x = x + \frac{1}{6} \ln | x | - \frac{9}{2} \ln | x - 2 | + \frac{28}{3} \ln | x - 3 | + C \end{matrix}$

(10) 最简分式：设 $\begin{matrix} (12) & 1 = A (x + 1) (x^{2} + 1) + B (x - 1) (x^{2} + 1) + E (x^{2} - 1) \end{matrix}$ 则 $\begin{matrix} (13) & \int \frac{1}{x^{4} - 1} d x = \int (\frac{A}{x - 1} + \frac{B}{x + 1} + \frac{E}{x^{2} + 1}) d x \end{matrix}$ 取 $x = i$ ，得到 $E = - \frac{1}{2}$ ；取 $x = 1$ ，得到 $A = \frac{1}{4}$ ；取 $x = - 1$ ，得到 $B = - \frac{1}{4}$ 或由偶函数知 $B = - A$ 。故 $\begin{matrix} (14) & \int \frac{1}{x^{4} - 1} d x = \frac{1}{4} \ln | \frac{x - 1}{x + 1} | - \frac{1}{2} \arctan x + C \end{matrix}$

(11) 设 $\begin{matrix} (15) & x^{4} = (x^{2} + 1) (x^{2} + 4) + A (x^{2} + 4) + B (x^{2} + 1) \end{matrix}$ 则 $\begin{matrix} (16) & \int \frac{x^{4}}{x^{4} + 5 x^{2} + 4} d x = \int (1 + \frac{A}{x^{2} + 1} + \frac{B}{x^{2} + 4}) d x \end{matrix}$ 取 $x = i$ 得到 $A = \frac{1}{3}$ ；取 $x = 2 i$ 得到 $B = - \frac{16}{3}$ 。故 $\begin{matrix} (17) & \int \frac{x^{4}}{x^{4} + 5 x^{2} + 4} d x = x + \frac{1}{3} \arctan x - \frac{8}{3} \arctan \frac{x}{2} + C \end{matrix}$

(12) 设 $\begin{matrix} (18) & 1 = A (x^{2} - x + 1) + [B (x - \frac{1}{2}) + E] (x + 1) \end{matrix}$ 则 $\begin{matrix} (19) & \int \frac{1}{x^{3} + 1} d x = \int [\frac{A}{x + 1} + \frac{B (x - \frac{1}{2}) + E}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}}] d x \end{matrix}$ 取 $x = - 1$ 得到 $A = \frac{1}{3}$ ；比较 $x^{2}$ 的系数得到 $B = - A = - \frac{1}{3}$ ；取 $x = 0$ 得到 $E = \frac{1}{2}$ 。故 $\begin{matrix} (20) & \int \frac{1}{x^{3} + 1} d x = \frac{1}{3} \ln | x + 1 | - \frac{1}{6} \ln (x^{2} - x + 1) + \frac{1}{\sqrt{3}} \arctan \frac{2 x - 1}{\sqrt{3}} + C \end{matrix}$

(13) $\begin{matrix} (21) & \begin{aligned} \int \frac{x^{7}}{{(1 - x^{2})}^{5}} d x & = - \frac{1}{2} \int \frac{(1 - y)^{3}}{y^{5}} d y, y = 1 - x^{2} \\ = \frac{4 x^{6} - 6 x^{4} + 4 x^{2} - 1}{8 (x^{2} - 1)^{4}} + C \end{aligned} \end{matrix}$

(14) 设 $\begin{matrix} (22) & 1 = (A x^{2} + B) (2 x^{2} - 1) + E x^{4} (\sqrt{2} x + 1) + F x^{4} (\sqrt{2} x - 1) \end{matrix}$ 则 $\begin{matrix} (23) & \int \frac{1}{x^{4} (2 x^{2} - 1)} d x = \int (\frac{A}{x^{2}} + \frac{B}{x^{4}} + \frac{E}{\sqrt{2} x - 1} + \frac{F}{\sqrt{2} x + 1}) d x \end{matrix}$ 取 $x = 0$ 得到 $B = - 1$ ；取 $x = \frac{1}{\sqrt{2}}$ 得到 $E = 2$ ；取 $x = - \frac{1}{\sqrt{2}}$ 得到， $F = - 2$ （或由偶函数知 $F = - E$ ）。求二阶导可得并代入 $x = 0$ 可得 $\begin{matrix} (24) & 0 = 2 A (2 x^{2} - 1) + 4 (A x^{2} + B) = - 2 A + 4 B ⟹ A = 2 B = - 2 \end{matrix}$

故 $\begin{matrix} (25) & \int \frac{1}{x^{4} (2 x^{2} - 1)} d x = \frac{2}{x} + \frac{1}{3 x^{3}} + \sqrt{2} \ln | \frac{\sqrt{2} x - 1}{\sqrt{2} x + 1} | + C \end{matrix}$

(15) $\begin{matrix} (26) & \begin{aligned} \int \frac{1}{x (x^{n} + a)} d x & = \int \frac{x^{n - 1}}{x^{n} (x^{n} + a)} d x = \frac{1}{n} \int \frac{d y}{y (y + a)}, y = x^{n} \\ = \frac{1}{n a} \ln | \frac{x^{n}}{x^{n} + a} | + C \end{aligned} \end{matrix}$ $◻$

8.2.2 三角函数的不定积分

例 8.2.2 计算以下不定积分：

(1): $\int (1 - 2 \cot^{2} x) d x$
(2): $\int \tan x d x$
(3): $\int \frac{\sec^{2} x}{\sqrt{1 + \tan x}} d x$
(4): $\int \cos^{2} (1 - 2 x) d x$
(5): $\int \cos^{3} x d x$
(6): $\int \sin α x \cos β x d x$
(7): $\int \tan^{4} x d x$
(8): $\int \sqrt{1 + \cos x} d x$
(9): $\int \frac{\sin 2 x}{1 + \sin^{4} x} d x$
(10): $\int \frac{\sin^{4} x}{\cos^{3} x} d x$
(11): $\int \frac{1}{\sin x \cos^{4} x} d x$
(12): $\int \frac{\sin^{2} x}{1 + \sin^{2} x} d x$
(13): $\int \frac{1 + \tan x}{\sin 2 x} d x$
(14): $\int \frac{1 - \tan x}{1 + \tan x} d x$
(15): $\int \frac{1}{(2 + \cos x) \sin x} d x$
(16): $\int \frac{\sin x}{\sin x + \cos x} d x$
(17): $\int \frac{1}{5 + 4 \sin x} d x$
(18): $\int \frac{\cos x}{\sin x + \cos x} d x$
(19): $\int \frac{\sin x \cos^{3} x}{1 + \cos^{2} x} d x$
(20): $\int \frac{\sqrt{1 + \cos x}}{\sin x} d x, x \in (0, π)$
(21): $\int \sqrt{1 + \csc x} d x$

解情况一：利用三角函数二倍角公式、积化和差等降次，用三角函数的平方关系化简。

(4) $\begin{matrix} (27) & \int \cos^{2} (1 - 2 x) d x = \int \frac{\cos (2 - 4 x) + 1}{2} d x = \frac{1}{2} x - \frac{1}{8} \sin (2 - 4 x) + C \end{matrix}$

(6) 设 $α^{2} \neq β^{2}$ ，则 $\begin{matrix} (28) & \begin{aligned} \int \sin α x \cos β x d x & = \int \frac{\sin (α + β) x + \sin (α - β) x}{2} d x \\ = - \frac{\cos (α + β) x}{2 (α + β)} - \frac{\cos (α - β) x}{2 (α - β)} + C \end{aligned} \end{matrix}$

(8) 设 $x \in [- π, π]$ ，则 $\begin{matrix} (29) & \int \sqrt{1 + \cos x} d x = \int \sqrt{2 \cos^{2} \frac{x}{2}} d x = 2 \sqrt{2} \int \cos \frac{x}{2} d \frac{x}{2} = 2 \sqrt{2} \sin \frac{x}{2} + C \end{matrix}$

情况二： $\int f (\tan x) d x, t = \tan x$ 。

(1) $\begin{matrix} (30) & \int (1 - 2 \cot^{2} x) d x = \int (1 - \frac{2}{t^{2}}) \frac{d t}{1 + t^{2}} = 3 x + 2 \cot x + C \end{matrix}$

(2) $\begin{matrix} (31) & \int \tan x d x = \int t \frac{d t}{1 + t^{2}} = \frac{1}{2} \ln (1 + \tan^{2} x) + C \end{matrix}$

(2') $\begin{matrix} (32) & \begin{aligned} \int \tan x d x & = - \int \frac{- \sin x d x}{\cos x} = - \ln | \cos x | + C \\ \int \tan x d x & = \int \frac{\sec x \tan x d x}{\sec x} = \ln | \sec x | + C \end{aligned} \end{matrix}$

(3) $\begin{matrix} (33) & \int \frac{\sec^{2} x}{\sqrt{1 + \tan x}} d x = \int \frac{1}{\sqrt{1 + t}} d (t + 1) = 2 \sqrt{1 + \tan x} + C \end{matrix}$

(7) $\begin{matrix} (34) & \int \tan^{4} x d x = \int t^{4} \frac{d t}{1 + t^{2}} = x - \tan x + \frac{1}{3} \tan^{3} x + C \end{matrix}$

(12) $\begin{matrix} (35) & \begin{aligned} \int \frac{\sin^{2} x}{1 + \sin^{2} x} d x & = \int \frac{\sin^{2} x}{2 \sin^{2} x + \cos^{2} x} d x = \int \frac{t^{2}}{2 t^{2} + 1} \frac{d t}{1 + t^{2}} \\ = x - \frac{1}{\sqrt{2}} \arctan (\sqrt{2} \tan x) + C \end{aligned} \end{matrix}$

(13) $\begin{matrix} (36) & \int \frac{1 + \tan x}{\sin 2 x} d x = \int \frac{1 + \tan x}{2 \tan x} d \tan x = \int \frac{1 + t}{2 t} d t = \frac{1}{2} \tan x + \frac{1}{2} \ln | \tan x | + C \end{matrix}$

(14) $\begin{matrix} (37) & \int \frac{1 - \tan x}{1 + \tan x} d x = \int \frac{1 - t}{1 + t} \frac{d t}{1 + t^{2}} = \ln | \cos x | + \ln | 1 + \tan x | + C \end{matrix}$

(14') $\begin{matrix} (38) & \int \frac{1 - \tan x}{1 + \tan x} d x = \int \frac{\cos x - \sin x}{\sin x + \cos x} d x = \ln | \sin x + \cos x | + C \end{matrix}$

(16) $\begin{matrix} (39) & \int \frac{\cos x}{\sin x + \cos x} d x = \int \frac{1}{\tan x + 1} d x = \int \frac{1}{1 + t} \frac{d t}{1 + t^{2}} = \frac{x}{2} + \frac{1}{2} \ln | \sin x + \cos x | + C \end{matrix}$

(18) $\begin{matrix} (40) & \int \frac{\sin x}{\sin x + \cos x} d x = \int \frac{\tan x}{\tan x + 1} d x = \int \frac{t}{1 + t} \frac{d t}{1 + t^{2}} = \frac{x}{2} - \frac{1}{2} \ln | \sin x + \cos x | + C \end{matrix}$

(16') (18') $\begin{matrix} (41) & \begin{aligned} (16) - (18) & = \int \frac{\cos x - \sin x}{\sin x + \cos x} d x = \ln | \sin x + \cos x | + C \\ (16) + (18) & = \int d x = x + C \end{aligned} \end{matrix}$

情况三： $\int f (\sin x) \cos x d x$ ， $\int f (\cos x) \sin x d x$ 。

(5) $\begin{matrix} (42) & \int \cos^{3} x d x = \int (1 - \sin^{2} x) d \sin x = \sin x - \frac{1}{3} \sin^{3} x + C \end{matrix}$

(9) 令 $t = \sin x$ ，则 $\begin{matrix} (43) & \int \frac{\sin 2 x}{1 + \sin^{4} x} d x = \int \frac{d (t^{2})}{1 + t^{4}} = \arctan (\sin^{2} x) + C \end{matrix}$

(10) 令 $t = \sin x$ ，则 $\begin{matrix} (44) & \begin{aligned} \int \frac{\sin^{4} x}{\cos^{3} x} d x & = \int \frac{\sin^{4} x}{\cos^{4} x} d \sin x = \int \frac{t^{4}}{{(1 - t^{2})}^{2}} d t \\ = \frac{3}{4} \ln \frac{1 - \sin x}{1 + \sin x} + \frac{3}{2} \sec x \tan x - \sin x \tan^{2} x + C \end{aligned} \end{matrix}$

(11) 令 $t = \cos x$ ，则 $\begin{matrix} (45) & \begin{aligned} \int \frac{1}{\sin x \cos^{4} x} d x & = - \int \frac{d \cos x}{\sin^{2} x \cos^{4} x} = \int \frac{- 1}{(1 - t^{2}) t^{4}} d t \\ = \ln | \tan \frac{x}{2} | + \sec x + \frac{1}{3} \sec^{3} x + C \end{aligned} \end{matrix}$

(15) 令 $t = \cos x$ ，则 $\begin{matrix} (46) & \int \frac{1}{(2 + \cos x) \sin x} d x = - \int \frac{d t}{(2 + t) (1 - t^{2})} = \frac{1}{6} \ln \frac{(1 - \cos x) (2 + \cos x)^{2}}{(1 + \cos x)^{3}} + C \end{matrix}$

(19) 令 $t = \cos x$ ，则 $\begin{matrix} (47) & \int \frac{\sin x \cos^{3} x}{1 + \cos^{2} x} d x = - \int \frac{t^{3}}{1 + t^{2}} d t = - \frac{1}{4} \cos 2 x + \frac{1}{2} \ln (3 + \cos 2 x) + C \end{matrix}$

(20) 令 $t = \cos x$ ，则 $\begin{matrix} (48) & \int \frac{\sqrt{1 + \cos x}}{\sin x} d x = - \int \frac{\sqrt{1 + t}}{1 - t^{2}} d t = \sqrt{2} \ln | \tan \frac{x}{4} | + C \end{matrix}$

情况四：万能公式。令 $t = \tan \frac{x}{2}$ ，则 $\cos x = \frac{1 - t^{2}}{1 + t^{2}}$ ， $\sin x = \frac{2 t}{1 + t^{2}}$ 。

(17) $\begin{matrix} (49) & \begin{aligned} \int \frac{1}{5 + 4 \sin x} d x & = \int \frac{1}{5 + 4 \frac{2 t}{1 + t^{2}}} d (2 \arctan t) = \int \frac{2}{5 + 5 t^{2} + 8 t} d t \\ = \frac{2}{3} \arctan \frac{4 + 5 \tan \frac{x}{2}}{3} + C \end{aligned} \end{matrix}$

(21) $\begin{matrix} (50) & \begin{aligned} \int \sqrt{1 + \csc x} d x & = \int \sqrt{1 + \frac{1 + t^{2}}{2 t}} d (2 \arctan t) = \int \frac{t + 1}{\sqrt{2 t}} \frac{2}{1 + t^{2}} d t \\ = - 2 \arctan \frac{\cot x}{\sqrt{1 + \csc x}} + C \end{aligned} \end{matrix}$ $◻$

8.2.3 无理式的不定积分

例 8.2.3 计算以下不定积分：

(1): $\int \frac{x^{2}}{\sqrt{a^{2} + x^{2}}} d x$
(2): $\int \frac{\sqrt{x^{2} - 4}}{x} d x$
(3): $\int \frac{1}{x \sqrt{a^{2} - x^{2}}} d x$
(4): $\int \frac{1}{x^{2} \sqrt{x^{2} - 1}} d x$
(5): $\int \frac{2 x - 1}{\sqrt{4 x^{2} + 4 x + 5}} d x$
(6): $\int \frac{x^{2}}{\sqrt{3 + 2 x - x^{2}}} d x$
(7): $\int \frac{1}{\sqrt{x} (\sqrt{x} + \sqrt[3]{x})} d x$
(8): $\int \frac{\sqrt{x + 1} - \sqrt{x - 1}}{\sqrt{x + 1} + \sqrt{x - 1}} d x$
(9): $\int x \sqrt{x + 2} d x$
(10): $\int x^{2} \sqrt{1 - x^{2}} d x$
(11): $\int x \sqrt{x^{4} + 2 x^{2} - 1} d x$
(12): $\int x \sqrt{\frac{1 + x}{1 - x}} d x$
(13): $\int \sqrt{\frac{a - x}{x - b}} d x$
(14): $\int \frac{1 - x + x^{2}}{\sqrt{1 + x - x^{2}}} d x$
(15): $\int \frac{1}{\sqrt{{(a^{2} - x^{2})}^{3}}} d x$

解 (1) 解法一：三角换元 $x = a \tan t$ 。设 $\begin{matrix} (51) & u^{2} = A (1 - u) (1 + u)^{2} + A (1 + u) (1 - u)^{2} + B (1 + u)^{2} + B (1 - u)^{2} \end{matrix}$ 则 $\begin{matrix} (52) & \begin{aligned} \int \frac{x^{2}}{\sqrt{a^{2} + x^{2}}} d x & = \int \frac{a^{2} \tan^{2} t}{\frac{a}{\cos t}} \frac{a}{\cos^{2} t} d t = a^{2} \int \frac{\sin^{2} t}{{(1 - \sin^{2} t)}^{2}} d \sin t \\ = a^{2} \int [\frac{A}{1 - u} + \frac{A}{1 + u} + \frac{B}{(1 - u)^{2}} + \frac{B}{(1 + u)^{2}}] d u, u = \sin t \end{aligned} \end{matrix}$ 取 $u = 1$ 得到 $B = \frac{1}{4}$ ，取 $u = 0$ 得到 $A = - B = - \frac{1}{4}$ ，所以 $\begin{matrix} (53) & \begin{aligned} \int \frac{x^{2}}{\sqrt{a^{2} + x^{2}}} d x & = \frac{a^{2}}{4} \ln \frac{1 - u}{1 + u} + \frac{a^{2}}{2} \frac{u}{1 - u^{2}} + C \\ = \frac{a^{2}}{4} \ln \frac{1 - \sin t}{1 + \sin t} + \frac{a^{2}}{2} \frac{\sin t}{\cos^{2} t} + C \\ = \frac{a^{2}}{4} \ln \frac{\sqrt{x^{2} + a^{2}} - x}{\sqrt{x^{2} + a^{2}} + x} + \frac{1}{2} x \sqrt{x^{2} + a^{2}} + C \\ = \frac{1}{2} x \sqrt{a^{2} + x^{2}} + \frac{a^{2}}{2} \ln (\sqrt{a^{2} + x^{2}} - x) + C \end{aligned} \end{matrix}$

解法二：双曲函数换元 $x = a \sinh t$ ，则 $\begin{matrix} (54) & \frac{x}{a} + \sqrt{1 + \frac{x^{2}}{a^{2}}} = e^{t}, \sqrt{1 + \frac{x^{2}}{a^{2}}} - \frac{x}{a} = e^{- t} \end{matrix}$ 故有 $\begin{matrix} (55) & \begin{aligned} \int \frac{x^{2}}{\sqrt{a^{2} + x^{2}}} d x & = \int \frac{a^{2} \sinh^{2} t}{a \cosh t} a \cosh t d t \\ = a^{2} \int {(\frac{e^{t} - e^{- t}}{2})}^{2} d t = \frac{a^{2}}{4} \int (e^{2 t} + e^{- 2 t} - 2) d t \\ = \frac{a^{2}}{4} [\frac{e^{2 t} - e^{- 2 t}}{2} - 2 t] + C \end{aligned} \end{matrix}$ 其中 $\begin{matrix} (56) & e^{2 t} - e^{- 2 t} = {(\frac{x}{a} + \sqrt{1 + \frac{x^{2}}{a^{2}}})}^{2} - {(\sqrt{1 + \frac{x^{2}}{a^{2}}} - \frac{x}{a})}^{2} = \frac{2}{a^{2}} x \sqrt{x^{2} + a^{2}} \end{matrix}$ $\begin{matrix} (57) & - t = \ln (\sqrt{x^{2} + a^{2}} - x) - \ln a \end{matrix}$

解法三：双曲有理换元 $x = \frac{2 a t}{1 - t^{2}}$ ，则 $\begin{matrix} (58) & t = \frac{- a + \sqrt{a^{2} + x^{2}}}{x}, \sqrt{a^{2} + x^{2}} = \sqrt{a^{2} + \frac{4 a^{2} t^{2}}{{(1 - t^{2})}^{2}}} = a \frac{1 + t^{2}}{1 - t^{2}} \end{matrix}$ 故有 $\begin{matrix} (59) & \int \frac{x^{2}}{\sqrt{a^{2} + x^{2}}} d x = \int \frac{4 a^{2} t^{2}}{{(1 - t^{2})}^{2}} \cdot \frac{1 - t^{2}}{a (1 + t^{2})} d \frac{2 a t}{1 - t^{2}} = 8 a^{2} \int \frac{t^{2}}{{(1 - t^{2})}^{3}} d t \end{matrix}$

(2) 令 $t = \sqrt{x^{2} - 4}$ ，则 $\begin{matrix} (60) & \begin{aligned} \int \frac{\sqrt{x^{2} - 4}}{x} d x & = \frac{1}{2} \int \frac{\sqrt{x^{2} - 4}}{x^{2}} d x^{2} = \frac{1}{2} \int \frac{t}{t^{2} + 4} d (t^{2} + 4) \\ = \sqrt{x^{2} - 4} - 2 \arctan \frac{\sqrt{x^{2} - 4}}{2} + C \end{aligned} \end{matrix}$

(3) 令 $t = \sqrt{a^{2} - x^{2}}$ ，则 $\begin{matrix} (61) & \int \frac{1}{x \sqrt{a^{2} - x^{2}}} d x = \frac{1}{2} \int \frac{d (a^{2} - t^{2})}{(a^{2} - t^{2}) t} = \frac{1}{2 a} \ln \frac{a - \sqrt{a^{2} - x^{2}}}{a + \sqrt{a^{2} - x^{2}}} + C \end{matrix}$

(4) 解法一：三角换元 $x = \sec t$ ， $0 \leq t \leq π$ 且 $t \neq \frac{π}{2}$ ，则 $\begin{matrix} (62) & \int \frac{1}{x^{2} \sqrt{x^{2} - 1}} d x = \int \frac{\cos^{2} t}{\tan t} d \frac{1}{\cos t} = \int \cos t d t = \sin t + C = \sqrt{1 - \frac{1}{x^{2}}} + C \end{matrix}$

解法二：双曲换元 $x = \cosh t$ ，则 $\begin{matrix} (63) & \begin{aligned} \int \frac{1}{\cosh^{2} t \sinh t} d \cosh t & = \int \frac{1}{\cosh^{2} t} d t = \int \frac{4}{e^{2 t} + e^{- 2 t} + 2} d t \\ = \int \frac{2}{u^{2} + 2 u + 1} d u, u = e^{2 t} \\ = - \frac{2}{u + 1} + C \end{aligned} \end{matrix}$

(5) $\begin{matrix} (64) & \int \frac{2 x - 1}{\sqrt{4 x^{2} + 4 x + 5}} d x = \frac{1}{2} \sqrt{4 x^{2} + 4 x + 5} + \ln (\sqrt{4 x^{2} + 4 x + 5} - 2 x - 1) + C \end{matrix}$

(6) $\begin{matrix} (65) & \int \frac{x^{2}}{\sqrt{- x^{2} + 2 x + 3}} d x = - \frac{1}{2} (x + 3) \sqrt{- x^{2} + 2 x + 3} - 6 \tan^{- 1} \frac{\sqrt{- x^{2} + 2 x + 3}}{x + 1} + C \end{matrix}$

(7) $\begin{matrix} (66) & \int \frac{1}{\sqrt{x} (\sqrt[3]{x} + \sqrt{x})} d x = 6 \ln (\sqrt[6]{x} + 1) + C \end{matrix}$

(8) $\begin{matrix} (67) & \int \frac{\sqrt{x + 1} - \sqrt{x - 1}}{\sqrt{x - 1} + \sqrt{x + 1}} d x = \frac{1}{2} [x^{2} - \sqrt{x - 1} \sqrt{x + 1} x - 2 \log (\sqrt{x - 1} - \sqrt{x + 1}) - 1] + C \end{matrix}$

(9) $\begin{matrix} (68) & \int x \sqrt{x + 2} d x = \frac{2}{15} (x + 2)^{3 / 2} (3 x - 4) + C \end{matrix}$

(10) $\begin{matrix} (69) & \int x^{2} \sqrt{1 - x^{2}} d x = \frac{1}{8} [x \sqrt{1 - x^{2}} (2 x^{2} - 1) - 2 \tan^{- 1} \frac{\sqrt{1 - x^{2}}}{x + 1}] + C \end{matrix}$

(11) $\begin{matrix} (70) & \int x \sqrt{x^{4} + 2 x^{2} - 1} d x = \frac{1}{4} (x^{2} + 1) \sqrt{x^{4} + 2 x^{2} - 1} - \tanh^{- 1} \frac{\sqrt{x^{4} + 2 x^{2} - 1}}{x^{2} + \sqrt{2} + 1} + C \end{matrix}$

(12) $\begin{matrix} (71) & \int x \sqrt{\frac{x + 1}{1 - x}} d x = \frac{1}{2} \sqrt{\frac{x + 1}{1 - x}} (x^{2} + x - 2) - 2 \tan^{- 1} \frac{\sqrt{x + 1}}{\sqrt{2} - \sqrt{1 - x}} + C \end{matrix}$

(13) $\begin{matrix} (72) & \int \sqrt{\frac{a - x}{x - b}} d x = \sqrt{(a - x) (x - b)} + (a - b) \tan^{- 1} \sqrt{\frac{x - b}{a - x}} + C \end{matrix}$

(14) $\begin{matrix} (73) & \int \frac{x^{2} - x + 1}{\sqrt{- x^{2} + x + 1}} d x = \frac{1}{4} [(1 - 2 x) \sqrt{- x^{2} + x + 1} + 11 \tan^{- 1} \frac{x}{\sqrt{- x^{2} + x + 1} - 1}] + C \end{matrix}$

(15) $\begin{matrix} (74) & \int \frac{1}{\sqrt{{(a^{2} - x^{2})}^{3}}} d x = \frac{x}{a^{2} \sqrt{a^{2} - x^{2}}} + C \end{matrix}$ $◻$

8.2.4 换元法和分部积分

例 8.2.4 计算以下不定积分：

(1): $\int \frac{1}{(1 + x^{2}) \arctan x} d x$
(2): $\int \frac{1}{x^{2}} \sinh \frac{1}{x} d x$
(3): $\int x \sec^{2} (1 - x^{2}) d x$
(4): $\int \frac{x}{\sqrt{1 + x^{2}}} \sin \sqrt{1 + x^{2}} d x$
(5): $\int \sqrt{\frac{\arcsin x}{1 - x^{2}}} d x$
(6): $\int \frac{2^{x}}{\sqrt{4 - 4^{x + 1}}} d x$
(7): $\int \frac{e^{x}}{1 + e^{2 x}} d x$
(8): $\int \tanh x d x$
(9): $\int \frac{1}{x \ln x \ln \ln x} d x$
(10): $\int \frac{\sqrt{1 - \ln x}}{x} d x$

解 (1) $\begin{matrix} (75) & \int \frac{1}{(1 + x^{2}) \arctan x} d x = \int \frac{1}{\arctan x} d \arctan x = \ln \arctan x + C \end{matrix}$

(2) $\begin{matrix} (76) & \int \frac{1}{x^{2}} \sinh \frac{1}{x} d x = - \int d \cosh \frac{1}{x} = - \cosh \frac{1}{x} + C \end{matrix}$

(3) $\begin{matrix} (77) & \int x \sec^{2} (1 - x^{2}) d x = - \frac{1}{2} \int \sec^{2} u d u = \tan (1 - x^{2}) + C \end{matrix}$

(4) $\begin{matrix} (78) & \int \frac{x}{\sqrt{1 + x^{2}}} \sin \sqrt{1 + x^{2}} d x = - \cos \sqrt{1 + x^{2}} + C \end{matrix}$

(5) $\begin{matrix} (79) & \int \sqrt{\frac{\arcsin x}{1 - x^{2}}} d x = \frac{2}{3} (\arcsin x)^{\frac{3}{2}} + C \end{matrix}$

(6) $\begin{matrix} (80) & \int \frac{2^{x}}{\sqrt{4 - 4^{x + 1}}} d x = \frac{1}{2} \int \frac{u}{\sqrt{1 - u^{2}}} d \frac{\ln u}{\ln 2} = \frac{1}{2 \sqrt{2}} \ln \arcsin (2^{x}) + C \end{matrix}$

(7) $\begin{matrix} (81) & \int \frac{e^{x}}{1 + e^{2 x}} d x = \arctan (e^{x}) + C \end{matrix}$

(8) $\begin{matrix} (82) & \int \tanh x d x = \int \frac{d \cosh x}{\cosh x} = \ln \cosh x + C \end{matrix}$

(9) $\begin{matrix} (83) & \int \frac{1}{x \ln x \ln \ln x} d x = \int \frac{1}{\ln x \ln \ln x} d \ln x = \ln \ln \ln x + C \end{matrix}$

(10) $\begin{matrix} (84) & \int \frac{\sqrt{1 - \ln x}}{x} d x = - \frac{2}{3} (1 - \ln x)^{\frac{3}{2}} + C \end{matrix}$ $◻$

例 8.2.5 计算以下不定积分：

(1): $\int x \cos 2 x d x$
(2): $\int x e^{- 3 x} d x$
(3): $\int x^{2} \sin^{2} x d x$
(4): $\int x \arctan x d x$
(5): $\int x \ln (x - 1) d x$
(6): $\int \ln (x + \sqrt{1 + x^{2}}) d x$
(7): $\int \arccos^{2} x d x$
(8): $\int x \tan^{2} x d x$
(9): $\int \frac{x}{\sin^{2} x} d x$
(10): $\int e^{x} \sin^{2} x d x$
(11): $\int \frac{\arcsin e^{x}}{e^{x}} d x$
(12): $\int \sin (\ln x) d x$

解 (1) $\begin{matrix} (85) & \int x \cos 2 x d x = (A_{1} x + A_{2}) \cos 2 x + (B_{1} x + B_{2}) \sin 2 x + C \end{matrix}$ 两边求导得到 $\begin{matrix} (86) & x \cos 2 x = A_{1} \cos 2 x - 2 (A_{1} x + A_{2}) \sin 2 x + B_{1} \sin 2 x + 2 (B_{1} x + B_{2}) \cos 2 x \end{matrix}$ 比较系数得到 $\begin{matrix} (87) & B_{1} = \frac{1}{2}, A_{1} = 0, B_{2} = 0, A_{2} = \frac{1}{4} \end{matrix}$ 因此 $\begin{matrix} (88) & \int x \cos 2 x d x = \frac{1}{2} x \sin 2 x + \frac{1}{4} \cos 2 x + C \end{matrix}$

(2) $\begin{matrix} (89) & \int x e^{- 3 x} d x = (A x + B) e^{- 3 x} + C \end{matrix}$ 两边求导得到 $\begin{matrix} (90) & x e^{- 3 x} = e^{- 3 x} (- 3 A x - 3 B + A) \end{matrix}$ 比较系数得到 $\begin{matrix} (91) & A = - \frac{1}{3}, B = - \frac{1}{9} \end{matrix}$ 所以 $\begin{matrix} (92) & \int x e^{- 3 x} d x = - \frac{3 x + 1}{9} e^{- 3 x} + C \end{matrix}$

(3) $\begin{matrix} (93) & \begin{aligned} \int x^{2} \sin^{2} x d x & = \int \frac{x^{2} (1 - \cos 2 x)}{2} d x \\ = \frac{x^{3}}{6} + (A_{1} x^{2} + A_{2} x + A_{3}) \cos 2 x + (B_{1} x^{2} + B_{2} x + B_{3}) \sin 2 x + C \end{aligned} \end{matrix}$ 求导得 $\begin{matrix} (94) & \begin{aligned} x^{2} \frac{1 - \cos 2 x}{2} = & \frac{x^{2}}{2} + (2 A_{1} x + A_{2} + 2 B_{1} x^{2} + 2 B_{2} x + 2 B_{3}) \cos 2 x \\ + (- 2 A_{1} x^{2} - 2 A_{2} x - 2 A_{3} + 2 B_{1} x + B_{2}) \sin 2 x \end{aligned} \end{matrix}$ 比较系数得到 $\begin{matrix} (95) & B_{1} = A_{2} = - \frac{1}{4}, B_{3} = \frac{1}{8} A_{1} = B_{2} = A_{3} = 0 \end{matrix}$ 因此 $\begin{matrix} (96) & \int x^{2} \sin^{2} x d x = \frac{x^{3}}{6} - \frac{1}{4} x \cos 2 x + (- \frac{1}{4} x^{2} + \frac{1}{8}) \sin 2 x + C \end{matrix}$

(4) $\begin{matrix} (97) & \begin{aligned} \int x \arctan x d x & = \frac{x^{2} \arctan x}{2} - \int \frac{x^{2}}{2 (1 + x^{2})} d x \\ = \frac{x^{2} \arctan x}{2} - \frac{x}{2} + \frac{1}{2} \arctan x + C \end{aligned} \end{matrix}$

(5) $\begin{matrix} (98) & \int x \ln (x - 1) d x = \frac{x^{2} \ln (x - 1)}{2} - \int \frac{x^{2}}{2 (x - 1)} d x \end{matrix}$

(6) 令 $x = \frac{e^{t} - e^{- t}}{2} = \sinh t$ ，则 $\begin{matrix} (99) & \begin{aligned} \int \ln (x + \sqrt{1 + x^{2}}) d x & = \int \ln (\frac{e^{t} - e^{- t}}{2} + \frac{e^{t} + e^{- t}}{2}) d \frac{e^{t} - e^{- t}}{2} \\ = \int t d \sinh t = t \sinh t - \cosh t + C \\ = x \ln (x + \sqrt{1 + x^{2}}) - \sqrt{1 + x^{2}} + C \end{aligned} \end{matrix}$

(7) $\begin{matrix} (100) & \int (\arccos x)^{2} d x = \int t^{2} d \cos t = (A_{1} t^{2} + A_{2} t + A_{3}) \cos t + (B_{1} t^{2} + B_{2} t + B_{3}) \sin t + C \end{matrix}$ 求导得 $\begin{matrix} (101) & - t^{2} \sin t = (2 A_{1} t + A_{2} + B_{1} t^{2} + B_{2} t + B_{3}) \cos t + (2 B_{1} t + B_{2} - A_{1} t^{2} - A_{2} t - A_{3}) \sin t \end{matrix}$ 比较系数得到 $\begin{matrix} (102) & B_{1} = A_{2} = B_{3} = 0, A_{1} = 1, B_{2} = A_{3} = - 2 \end{matrix}$ 因此 $\begin{matrix} (103) & \begin{aligned} \int (\arccos x)^{2} d x & = (t^{2} - 2) \cos t - 2 t \sin t + C \\ = [(\arccos x)^{2} - 2] x - 2 \sqrt{1 - x^{2}} \arccos x + C \end{aligned} \end{matrix}$

(8) $\begin{matrix} (104) & \int x \tan^{2} x d x = - \frac{x^{2}}{2} + x \tan x + \ln | \cos x | + C \end{matrix}$

(9) $\begin{matrix} (105) & \int \frac{x}{\sin^{2} (x)} d x = \ln | \sin x | - x \cot x + C \end{matrix}$

(10) $\begin{matrix} (106) & \int e^{x} \sin^{2} x d x = e^{x} (A_{1} + A_{2} \sin 2 x + A_{3} \cos 2 x) + C \end{matrix}$ 求导得到 $\begin{matrix} (107) & e^{x} \sin^{2} x = \frac{e^{x} (1 - \cos 2 x)}{2} = e^{x} (A_{1} + A_{2} \sin 2 x + A_{3} \cos 2 x + 2 A_{2} \cos 2 x - 2 A_{3} \sin 2 x) \end{matrix}$ 比较系数得到 $\begin{matrix} (108) & A_{1} = \frac{1}{2}, A_{2} - 2 A_{3} = 0, A_{3} + 2 A_{2} = - \frac{1}{2} \end{matrix}$ 因此 $\begin{matrix} (109) & A_{1} = \frac{1}{2}, A_{2} = - \frac{1}{5}, A_{3} = - \frac{1}{10} \end{matrix}$ 亦即 $\begin{matrix} (110) & \int e^{x} \sin^{2} x d x = \frac{1}{2} e^{x} - \frac{1}{5} e^{x} \sin 2 x - \frac{1}{10} e^{x} \cos 2 x + C \end{matrix}$

(11) $\begin{matrix} (111) & \int \frac{\sin^{- 1} e^{x}}{e^{x}} d x = - e^{- x} \sin^{- 1} e^{x} - \tanh^{- 1} \sqrt{1 - e^{2 x}} + C \end{matrix}$

(12) $\begin{matrix} (112) & \int \sin (\ln x) d x = - \frac{1}{2} x [\cos (\ln x) - \sin (\ln x)] + C \end{matrix}$ $◻$

8.2.5 杂题

例 8.2.6 以下函数是否存在原函数？若存在，求它的不定积分。

(1): $f (x) = | (x - 1) (3 x - 2) |$
(2): $sgn (x) = {\begin{cases} 1, & x > 0, \\ 0, & x = 0, \\ - 1, & x < 0 . \end{cases}$
(3): $f (x) = {\begin{cases} - \cos \frac{1}{x} + 2 x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0 . \end{cases}$

解 (1) 存在，设 $\begin{matrix} (113) & F (x) := C + {\begin{cases} 2 x - \frac{5}{2} x^{2} + x^{3}, & x \in (- \infty, \frac{2}{3}] \\ - 2 x + \frac{5}{2} x^{2} - x^{3} + \frac{28}{27}, & x \in (\frac{2}{3}, 1) \\ 2 x - \frac{5}{2} x^{2} + x^{3} + \frac{1}{27}, & x \in [1, + \infty) \end{cases} \end{matrix}$ 验证可知 $F^{'} (x) = f (x)$ 。

(2) 不存在，因为导函数不存在第一类间断点。

(3) 存在，设 $\begin{matrix} (114) & F (x) := C + {\begin{cases} x^{2} \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases} \end{matrix}$ 验证可知 $F^{'} (x) = f (x)$ 。 $◻$