求下面两道不定积分的详解
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/11/18 05:03:43
求下面两道不定积分的详解
(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