自修不定积分：吴传生《经济数学微积分》第四版总习题五（一）

最新推荐文章于 2025-05-27 18:19:13 发布

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最新推荐文章于 2025-05-27 18:19:13 发布

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分类专栏：数学文章标签：线性代数

本文链接：https://blog.youkuaiyun.com/qq_49547348/article/details/127920261

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前言

该章节的小节题目比较简单，但综合题目（总习题五）有一定难度。作为自修不定积分的学习笔记，此篇博文解析习题步骤，记录学习成果。

另外，本人只借助此书学习微积分相关知识，不希求学习经济数学专业内容，因此只做了第一大题，第二大题没有做。

录入匆忙，难免有纰漏，敬请指正。

正文开始

1、求下列不定积分（a,b为常数）
$\begin{aligned} (1)\int\frac{\mathrm{d}x}{\mathrm{e}^x+\mathrm{e}^{-x}} &=\int\frac{\mathrm{e}^x\mathrm{d}x}{\left(\mathrm{e}^x\right)^2+1}\\ &=\int\frac{\mathrm{d}\mathrm{e}^x}{\left(\mathrm{e}^x\right)^2+1}\\ &=\left(\int\frac{\mathrm{d}u}{u^2+1}\right)_{u=\mathrm{e}^x}\\ &=\arctan{u}+C\\ &=\arctan{\mathrm{e}^x}+C \end{aligned}$

$\begin{aligned} (2)\int\frac{1+x}{(1-x)^3}\mathrm{d}x &=\int\frac{2-1+x}{(1-x)^3}\mathrm{d}x\\ &=\left(\int\frac{2+u}{-u^3}\mathrm{d}u\right)_{u=x-1}\\ &=-2\int{u^{-3}\mathrm{d}u}-\int{u^{-2}\mathrm{d}u}\\ &=u^{-2}-u^{-1}+C\\ &=(x-1)^{-2}+(x-1)^{-1}+C \end{aligned}$

$\begin{aligned} (3)\int\frac{x^2}{1-x^6}\mathrm{d}x &=\int\frac{\mathrm{d}\frac{1}{3}x^3}{1-x^6}\\ &=\frac{1}{3}\int\frac{\mathrm{d}x^3}{1-x^6}\\ &=\left(\frac{1}{3}\int\frac{\mathrm{d}u}{1-u^2}\right)_{u=x^3}\\ &=-\frac{1}{3}\int\frac{\mathrm{d}u}{(u+1)(u-1)}\\ &=-\frac{1}{3}\int\frac{1}{2}\left(\frac{1}{u-1}-\frac{1}{u+1}\right)\mathrm{d}u\\ &=-\frac{1}{6}\int\frac{\mathrm{d}u}{u-1}+\frac{1}{6}\int\frac{\mathrm{d}u}{u+1}\\ &=-\frac{1}{6}\ln\left|u-1\right|+\frac{1}{6}\ln\left|u+1\right|+C\\ &=\frac{1}{6}\ln\left|\frac{u+1}{u-1}\right|+C\\ &=\frac{1}{6}\ln\left|\frac{x^3+1}{x^3-1}\right|+C \end{aligned}$

$\begin{aligned} (4)\int\frac{1-\cos{x}}{x-\sin{x}}\mathrm{d}x &=\int\frac{\mathrm{d}(x-\sin{x})}{x-\sin{x}}\\ &=\ln\left|x-\sin{x}\right|+C \end{aligned}$

$\begin{aligned} (5)\int\frac{\ln\ln{x}}{x}\mathrm{d}x &=\int\ln\ln{x}\frac{\mathrm{d}x}{x}\\ &=\int\ln\ln{x}\mathrm{d}\ln x\\ &=\left(\int\ln{u}\mathrm{d}u\right)_{u=\ln{x}}\\ &=u\ln{u}-\int{u}\mathrm{d}\ln{u}\\ &=u\ln{u}-\int{1}\mathrm{d}u\\ &=u\ln{u}-u+C\\ &=u(\ln{u}-1)+C\\ &=\ln{x}(\ln\ln{x}-1)+C \end{aligned}$

$\begin{aligned} (6)\int\frac{\mathrm{d}x}{x\left(x^9+1\right)} &=\int\frac{x^8\mathrm{d}x}{x^9\left(x^9+1\right)}\\ &=\frac{1}{9}\int\frac{\mathrm{d}x^9}{x^9\left(x^9+1\right)}\\ &=\frac{1}{9}\int\frac{1}{x^9}-\frac{1}{x^9+1}\mathrm{d}x^9\\ &=\frac{1}{9}\int\frac{\mathrm{d}x^9}{x^9}-\frac{1}{9}\int\frac{\mathrm{d}x^9}{x^9+1}\\ &=\frac{1}{9}\ln\left|x^9\right|-\frac{1}{9}\ln\left|x^9+1\right|+C\\ &=\frac{1}{9}\ln\left|\frac{x^9}{x^9+1}\right|+C \end{aligned}$