x^2/（1+x^2）^2的积分是多少,怎么算
展开全部
∫ x^2/(1+x^2)^2 dx
= ∫ dx/(1+x^2) -∫ dx/(1+x^2)^2
=arctanx -∫ dx/(1+x^2)^2
let
x=tany
dx = (secy)^2 dy
∫ dx/(1+x^2)^2
=∫ (cosy)^2 dy
=(1/2)∫ (1+cos2y) dy
=(1/2)[ y+(1/2)sin(2y) ]
=(1/2)[ arctanx +( x/(1+x^2) ) ]
∫ x^2/(1+x^2)^2 dx
=arctanx -∫ dx/(1+x^2)^2
=arctanx -(1/2)[ arctanx +(x/(1+x^2)) ]
=(1/2)arctanx -(1/2)[x/(1+x^2)] + C
= ∫ dx/(1+x^2) -∫ dx/(1+x^2)^2
=arctanx -∫ dx/(1+x^2)^2
let
x=tany
dx = (secy)^2 dy
∫ dx/(1+x^2)^2
=∫ (cosy)^2 dy
=(1/2)∫ (1+cos2y) dy
=(1/2)[ y+(1/2)sin(2y) ]
=(1/2)[ arctanx +( x/(1+x^2) ) ]
∫ x^2/(1+x^2)^2 dx
=arctanx -∫ dx/(1+x^2)^2
=arctanx -(1/2)[ arctanx +(x/(1+x^2)) ]
=(1/2)arctanx -(1/2)[x/(1+x^2)] + C
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
