求下面两道不定积分的详解
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/18 08:47:26
求下面两道不定积分的详解
![](http://img.wesiedu.com/upload/9/af/9af07c591928ced8947d38f479f6a5aa.jpg)
![](http://img.wesiedu.com/upload/9/af/9af07c591928ced8947d38f479f6a5aa.jpg)
![求下面两道不定积分的详解](/uploads/image/z/18194622-6-2.jpg?t=%E6%B1%82%E4%B8%8B%E9%9D%A2%E4%B8%A4%E9%81%93%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E7%9A%84%E8%AF%A6%E8%A7%A3)
(1)∫[xe^x/(e^x-1)^2]dx=∫xd(e^x-1)/(e^x-1)^2=-∫xd[1/(e^x-1)]=-xln(e^x-1)+∫dx/(e^x-1)
=-xln(e^x-1)+ln(e^x-1)-x+C=[(1-x)ln(e^x-1)]-x+C
(2)∫[1/(1-x)^2]*ln[(1+x)/(1-x)] dx=∫[1/(1-x^2)]dx*ln[(1+x)/(1-x)]=∫(1/2)[ln[(1+x)/(1-x)]d[ln[(1+x)/(1-x)]
=[ln[(1+x)/(1-x)]^2 /4+C
=-xln(e^x-1)+ln(e^x-1)-x+C=[(1-x)ln(e^x-1)]-x+C
(2)∫[1/(1-x)^2]*ln[(1+x)/(1-x)] dx=∫[1/(1-x^2)]dx*ln[(1+x)/(1-x)]=∫(1/2)[ln[(1+x)/(1-x)]d[ln[(1+x)/(1-x)]
=[ln[(1+x)/(1-x)]^2 /4+C