第五章 不定积分

第五章 不定积分#

课后习题解答#

5.1 不定积分的概念与性质#

  1. 设 \(f'(x) = \sin x\), 求 \(f(x)\) 的全体原函数.

解

由不定积分的定义, 有

\[f(x) = \int \sin x ~ \mathrm{d}x = -\cos x + C.\]

那么, \(f(x)\) 的全体原函数, 即它的不定积分为

\[\int f(x) ~ \mathrm{d}x = \int \left( -\cos x + C \right) ~ \mathrm{d}x = -\sin x + Cx + D,\]

其中 \(C, D\) 为任意常数.

备注

这题是已知函数的导函数, 求它的原函数, 不是求该函数本身, 注意文字上的细微差别.

§5.2 换元积分法#

§5.3 分部积分法#

  1. 已知 \(f(x) = \frac{e^x}{x}\), 求 \(\displaystyle \int x f''(x) ~ \mathrm{d}x\).

解

利用分部积分法, 有

\[\begin{split}\int x f''(x) ~ \mathrm{d}x & = \int x ~ \mathrm{d} f'(x) = x f'(x) - \int f'(x) ~ \mathrm{d}x \\ & = x f'(x) - f(x) + C.\end{split}\]

将 \(f(x) = \frac{e^x}{x}\) 代入上式, 得

\[\int x f''(x) ~ \mathrm{d}x = x \left( \frac{e^x}{x} \right)' - \frac{e^x}{x} + C = x \cdot \frac{(x - 1) e^x}{x^2} - \frac{e^x}{x} + C = \frac{(x - 2) e^x}{x} + C.\]

§5.4 几种特殊类型函数的不定积分#

  1. 试不用万用转换公式计算 \(\displaystyle \int \dfrac{1}{\sin^4 x} ~ \mathrm{d}x\).

解

注意到 \(\dfrac{1}{\sin^4 x} = \csc^4 x\), 因此有

\[\begin{split}\int \dfrac{1}{\sin^4 x} ~ \mathrm{d}x & = \int \csc^4 x ~ \mathrm{d}x = \int \csc^2 x \cdot \csc^2 x ~ \mathrm{d}x \\ & = - \int (1 + \cot^2 x) ~ \mathrm{d} (\cot x) \\ & = - \cot x - \dfrac{1}{3} \cot^3 x + C.\end{split}\]

补充内容#

  1. 求不定积分 \(\displaystyle \int \dfrac{\sqrt{1-x} \arctan \sqrt{1-x}}{2 - x} \mathrm{d}x\).

解

令 \(t = \sqrt{1-x}\), 那么 \(x = 1 - t^2\), 于是

\[\begin{split}\int \dfrac{\sqrt{1-x} \arctan \sqrt{1-x}}{2 - x} \mathrm{d}x & = \int \dfrac{t \arctan t}{1 + t^2} (-2t) \mathrm{d}t \\ & = -2 \int \dfrac{t^2 \arctan t}{1 + t^2} \mathrm{d}t \\ & = -2 \int \left( 1 - \dfrac{1}{1 + t^2} \right) \arctan t \mathrm{d}t \\ & = 2 \int \arctan t \mathrm d (\arctan t) - 2 \int \arctan t \mathrm{d}t \\ & = \arctan^2 t - 2 \int \arctan t \mathrm{d}t \\ & = \arctan^2 t -2 t \arctan t + 2 \int \dfrac{t}{1 + t^2} \mathrm{d}t \\ & = \arctan^2 t -2 t \arctan t + \ln(1 + t^2) + C\end{split}\]

回代 \(t = \sqrt{1-x}\), 得

\[\int \dfrac{\sqrt{1-x} \arctan \sqrt{1-x}}{2 - x} \mathrm{d}x = \arctan^2 \sqrt{1-x} -2 \sqrt{1-x} \arctan \sqrt{1-x} + \ln(2 - x) + C.\]

  1. 求不定积分

    (1). \(\displaystyle \int x^2 \sqrt{x} ~ \mathrm{d}x\)

    (2). \(\displaystyle \int \dfrac{1}{x^4 \sqrt{x^2 + 1}} ~ \mathrm{d}x\)

    (3). \(\displaystyle \int \dfrac{1}{\sin^2 x \cos^2 x} ~ \mathrm{d}x\)

    (4). \(\displaystyle \int \dfrac{1}{x^2 - 8x + 25} ~ \mathrm{d}x\)

    (5). \(\displaystyle \int \dfrac{x^5}{\sqrt{1 + x^2}} ~ \mathrm{d}x\)

    (6). \(\displaystyle \int x^3 \ln x ~ \mathrm{d}x\)

解

(1). 做变量替换 \(t = \sqrt{x}\), 那么 \(x = t^2\), \(\mathrm{d}x = 2t \mathrm{d}t\), 从而有

\[\int x^2 \sqrt{x} ~ \mathrm{d}x = \int t^4 \cdot t \cdot 2t ~ \mathrm{d}t = 2 \int t^6 ~ \mathrm{d}t = \dfrac{2}{7} t^7 + C = \dfrac{2}{7} x^{\frac{7}{2}} + C\]

这题也可以直接做:

\[\int x^2 \sqrt{x} ~ \mathrm{d}x = \int x^{\frac{5}{2}} ~ \mathrm{d}x = \dfrac{2}{7} x^{\frac{7}{2}} + C\]

(2). 做变量替换 \(t = 1/x\), 那么有

\[\begin{split}\int \dfrac{1}{x^4 \sqrt{x^2 + 1}} ~ \mathrm{d}x & = -\int \dfrac{t^4}{\sqrt{\frac{1}{t^2} + 1}} \cdot \dfrac{1}{t^2} ~ \mathrm{d}t \\ & = -\dfrac{1}{2} \int \dfrac{t^2 + 1 - 1}{\sqrt{t^2 + 1}} ~ \mathrm{d}(t^2 + 1) \\ & = -\dfrac{1}{2} \int \sqrt{t^2 + 1} ~ \mathrm{d}(t^2 + 1) + \dfrac{1}{2} \int \dfrac{1}{\sqrt{t^2 + 1}} ~ \mathrm{d}(t^2 + 1) \\ & = -\dfrac{1}{3} (t^2 + 1)^{\frac{3}{2}} + \sqrt{t^2 + 1} + C \\ & = -\dfrac{1}{3} \left( \dfrac{1}{x^2} + 1 \right)^{\frac{3}{2}} + \sqrt{\dfrac{1}{x^2} + 1} + C \\ & = -\dfrac{(x^2 + 1) \sqrt{x^2 + 1}}{3x^3} + \dfrac{\sqrt{x^2 + 1}}{x} + C \\ & = \dfrac{2x^2 - 1}{3x^3} \sqrt{x^2 + 1} + C\end{split}\]

(3). 注意到 \(1 = \sin^2 x + \cos^2 x\), 从而有

\[\begin{split}\int \dfrac{1}{\sin^2 x \cos^2 x} ~ \mathrm{d}x & = \int \dfrac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} ~ \mathrm{d}x = \int \left( \dfrac{1}{\sin^2 x} + \dfrac{1}{\cos^2 x} \right) ~ \mathrm{d}x \\ & = - \cot x + \tan x + C\end{split}\]

(4).

\[\int \dfrac{1}{x^2 - 8x + 25} ~ \mathrm{d}x = \int \dfrac{1}{(x - 4)^2 + 3^2} ~ \mathrm{d}(x - 4) = \dfrac{1}{3} \arctan \dfrac{x - 4}{3} + C\]

(5).

\[\begin{split}\int \dfrac{x^5}{\sqrt{1 + x^2}} ~ \mathrm{d}x & = \dfrac{1}{2} \int \dfrac{x^4 ~ \mathrm{d}(x^2 + 1)}{\sqrt{x^2 + 1}} \\ & = \dfrac{1}{2} \int \dfrac{(x^2 + 1 - 1)^2 ~ \mathrm{d}(x^2 + 1)}{\sqrt{x^2 + 1}} \\ & = \dfrac{1}{2} \int \left( (x^2 + 1)^{\frac{3}{2}} - 2 (x^2 + 1)^{\frac{1}{2}} + (x^2 + 1)^{-\frac{1}{2}} \right) ~ \mathrm{d}(x^2 + 1) \\ & = \dfrac{1}{2} \left( \dfrac{2}{5} (x^2 + 1)^{\frac{5}{2}} - \dfrac{4}{3} (x^2 + 1)^{\frac{3}{2}} + 2 \sqrt{x^2 + 1} \right) + C \\ & = \dfrac{3 x^4 - 4 x^2 + 8}{15} \sqrt{x^2 + 1} + C\end{split}\]

(6).

\[\begin{split}\int x^3 \ln x ~ \mathrm{d}x & = \dfrac{1}{4} \int \ln x ~ \mathrm{d}(x^4) = \dfrac{1}{4} x^4 \ln x - \dfrac{1}{4} \int x^3 ~ \mathrm{d}x \\ & = \dfrac{1}{4} x^4 \ln x - \dfrac{1}{16} x^4 + C\end{split}\]
目录