doc2x识别试卷、数学公式

一个低调的工具，doc2x。不要被官网朴实的页面所迷惑，它非常强大。

微积分教材随机识别-正确率100%

选择的人大版微积分第三版214、215页。

= 3 \sqrt[3]{x} - 6 \sqrt[6]{x} + 6 \ln | \sqrt[6]{x} + 1 | + C

例 9 求不定积分 $\int \frac{1}{\sqrt{1 + e^{x}}} d x$

解: 令 $t = \sqrt{1 + e^{x}}$ ,则 $e^{x} = t^{2} - 1, x = \ln (t^{2} - 1), d x = \frac{2 t d t}{t^{2} - 1}$

所以 $\int \frac{1}{\sqrt{1 + e^{x}}} d x = \int \frac{2}{t^{2} - 1} d t = \int (\frac{1}{t - 1} - \frac{1}{t + 1}) d t$

= \ln | \frac{t - 1}{t + 1} | + C = \ln | \frac{\sqrt{1 + e^{x}} - 1}{\sqrt{1 + e^{x}} + 1} | + C

= \ln \frac{{(\sqrt{1 + e^{x}} - 1)}^{2}}{e^{x}} + C

= 2 \ln (\sqrt{1 + e^{x}} - 1) - \ln e^{x} + C

= 2 \ln (\sqrt{1 + e^{x}} - 1) - x + C

例 10 求不定积分 $\int \sqrt{a^{2} - x^{2}} d x (a > 0)$

解: 设 $x = a \sin t (t \in [- \frac{π}{2}, \frac{π}{2}])$ ,则 $d x = a \cos t d t$

\sqrt{a^{2} - x^{2}} = \sqrt{a^{2} - a^{2} \sin^{2} t} = a \cos t

于是 $\int \sqrt{a^{2} - x^{2}} d x = \int a \cos t \cdot a \cos t d t = a^{2} \int \cos^{2} t d t$

= a^{2} \int \frac{1 + \cos 2 t}{2} d t = \frac{a^{2}}{2} [t + \frac{1}{2} \sin 2 t] + C

= \frac{a^{2}}{2} [t + \sin t \cdot \cos t] + C

= \frac{a^{2}}{2} [t + \sin t \cdot \sqrt{1 - \sin^{2} t}] + C

= \frac{a^{2}}{2} [\frac{x}{a} \cdot \sqrt{1 - {(\frac{x}{a})}^{2}} + \arcsin \frac{x}{a}] + C

\begin{matrix} (5.24) & = \frac{x}{2} \cdot \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \arcsin \frac{x}{a} + C \end{matrix}

例 11 求不定积分 $\int \frac{d x}{\sqrt{x^{2} + a^{2}}} (a > 0)$

解: 设 $x = a \tan t (t \in (- \frac{π}{2}, \frac{π}{2}))$ ,则 $d x = a \sec^{2} t d t$

\sqrt{x^{2} + a^{2}} = \sqrt{a^{2} \tan^{2} t + a^{2}} = a \sec t

于是 $\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = \int \frac{a \sec^{2} t}{a \sec t} d t = \int \sec t d t$

= \ln | \sec t + \tan t | + C_{1}

= \ln | \sqrt{1 + \tan^{2} t} + \tan t | + C_{1}

= \ln | \frac{x}{a} + \sqrt{{(\frac{x}{a})}^{2} + 1} | + C_{1}

= \ln | x + \sqrt{x^{2} + a^{2}} | + C (其中, C = C_{1} - \ln a)

例 12 求不定积分 $\int \frac{d x}{\sqrt{x^{2} - a^{2}}} (a > 0)$

解: 令 $x = a \sec t (t \in (0, \frac{π}{2}) \cup (\frac{π}{2}, π))$ ,则 $d x = a \cdot \sec t \cdot \tan t \cdot d t$

\sqrt{x^{2} - a^{2}} = \sqrt{a^{2} \sec^{2} t - a^{2}} = a \tan t

于是 $\int \frac{d x}{\sqrt{x^{2} - a^{2}}} = \int \frac{1}{a \tan t} \cdot a \cdot \sec t \cdot \tan t \cdot d t$

= \int \sec t d t = \ln | \sec t + \tan t | + C_{1}

= \ln | \sec t + \sqrt{\sec^{2} t - 1} | + C_{1}

= \ln | \frac{x}{a} + \sqrt{{(\frac{x}{a})}^{2} - 1} | + C_{1}

= \ln | x + \sqrt{x^{2} - a^{2}} | + C (其中, C = C_{1} - \ln a)

合并例 11 、例 12 , 得

\begin{matrix} (5.25) & \int \frac{d x}{\sqrt{x^{2} \pm a^{2}}} = \ln | x + \sqrt{x^{2} \pm a^{2}} | + C (a > 0) \end{matrix}

如果 $u = u (x)$ 与 $v = v (x)$ 都有连续的导数,则由函数乘积的微分公式 $d (u v) = v d u + u d v$ 移项得

u d v = d (u v) - v d u