数学积分问题求过程∫dx/((x-a)(x-b))=1/(a-b)ln((x-a)/(x-b))+C

网友回答

不妨设a>b1/(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)|
======以下答案可供参考======
供参考答案1:
∫dx/((x-a)(x-b))
=∫(1/(a-b))((1/(x-a))-(1/(x-b)))dx
=(1/(a-b))(∫(1/(x-a))dx-∫(1/(x-b))dx)
=(1/(a-b))(ln(x-a)-ln(x-b))+C
=1/(a-b)ln((x-a)/(x-b))+C
供参考答案2:
1/(x-a)(x-b)
=[1/(a-b)]*[1/(x-a)-1/(x-b)]所以∫dx/((x-a)(x-b))
=[1/(a-b)]∫dx/[1/(x-a)-1/(x-b)]
=1/(a-b)ln((x-a)/(x-b))+C