![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
2个回答
展开全部
(3) 原式 = ∫(1/2)d(x^2)/√(a^2+x^2)
= ∫(1/2)d(x^2+a^2)/√(a^2+x^2)
= (1/2) 2√(a^2+x^2) + C = √(a^2+x^2) + C
= ∫(1/2)d(x^2+a^2)/√(a^2+x^2)
= (1/2) 2√(a^2+x^2) + C = √(a^2+x^2) + C
2017-12-16 · 知道合伙人教育行家
关注
![](https://wyw-base.cdn.bcebos.com/pc-content/follow.gif)
展开全部
4
∫[x/(1+x^2)]dx
=∫{1/[2(1+x^2)]}d(1+x^2)
=(1/2)*log|1+x^2|+c
2
y=√x,x=y^2
∫dx/[√x(1+x)]
=∫2ydy/[y(1+y^2)]
=∫2dy/(1+y^2)
=2arctany+c
=2arctan(√x)+c
1
y=√(1+2x),x=(y^2-1)/2
∫dx/√(1+2x)
=∫ydy/y
=y+c
=√(1+2x)+c
∫[x/(1+x^2)]dx
=∫{1/[2(1+x^2)]}d(1+x^2)
=(1/2)*log|1+x^2|+c
2
y=√x,x=y^2
∫dx/[√x(1+x)]
=∫2ydy/[y(1+y^2)]
=∫2dy/(1+y^2)
=2arctany+c
=2arctan(√x)+c
1
y=√(1+2x),x=(y^2-1)/2
∫dx/√(1+2x)
=∫ydy/y
=y+c
=√(1+2x)+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