数学积分问题求过程∫dx/((x-a)(x-b))=1/(a-b)ln((x-a)/(x-b))+Ca和b是常数且不相等_数学解答

数学积分问题求过程∫dx/((x-a)(x-b))=1/(a-b)ln((x-a)/(x-b))+C
a和b是常数且不相等

人气：412 ℃　时间：2020-05-01 11:02:33

解答

不妨设a>b
1/(x-a)(x-b)
=[1/(a-b)]*(a-b)/(x-a)(x-b)
=[1/(a-b)]*[(x-b)-(x-a)]/(x-a)(x-b)
=[1/(a-b)]*[(x-b)/(x-a)(x-b)-(x-a)/(x-a)(x-b)]
=[1/(a-b)]*[1/(x-a)-1/(x-b)]
所以原式=[1/(a-b)]*∫[1/(x-a)-1/(x-b)]dx
=[1/(a-b)]*[ln|(x-a)|-ln|x-b|]+C
=[1/(a-b)]*[ln|(x-a)/(x-b)|