问题∫[1/(x-2)(x+2)]dx=1/4∫[1/(x-2)-1/(x+2)]dx 是怎么得来的?原理?

来源:学生作业帮助网 编辑:六六作业网 时间:2024/05/26 01:42:32
问题∫[1/(x-2)(x+2)]dx=1/4∫[1/(x-2)-1/(x+2)]dx是怎么得来的?原理?问题∫[1/(x-2)(x+2)]dx=1/4∫[1/(x-2)-1/(x+2)]dx是怎么得

问题∫[1/(x-2)(x+2)]dx=1/4∫[1/(x-2)-1/(x+2)]dx 是怎么得来的?原理?
问题∫[1/(x-2)(x+2)]dx=1/4∫[1/(x-2)-1/(x+2)]dx 是怎么得来的?原理?

问题∫[1/(x-2)(x+2)]dx=1/4∫[1/(x-2)-1/(x+2)]dx 是怎么得来的?原理?
噢,假设1/(x-2)(x+2)=A/(x-2)+B/(x+2)
通分,比较系数可得A=1/4,B=-1/4