不定积分解答步骤
展开全部
∫ arctan√x dx
=x.arctan√x - (1/2)∫ √x/(1+x) dx
=x.arctan√x - ( √x - arctan√x ) + C
=2x.arctan√x - √x + C
----
let
u=√x
2udu = dx
∫ √x/(1+x) dx
=∫ [u/(1+u^2) ] (2udu)
=2∫ u^2/(1+u^2) du
=2∫[ 1- 1/(1+u^2) ]du
=2( u - arctanu ) + C'
=2( √x - arctan√x ) + C'
=x.arctan√x - (1/2)∫ √x/(1+x) dx
=x.arctan√x - ( √x - arctan√x ) + C
=2x.arctan√x - √x + C
----
let
u=√x
2udu = dx
∫ √x/(1+x) dx
=∫ [u/(1+u^2) ] (2udu)
=2∫ u^2/(1+u^2) du
=2∫[ 1- 1/(1+u^2) ]du
=2( u - arctanu ) + C'
=2( √x - arctan√x ) + C'
本回答由提问者推荐
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
