
2个回答
展开全部
下面的不定积分帮你求,上面那题是trivial的
∫ 1/(1+3sqrt(x))dx=
x=u^2, du=2udx
=1/2∫ u/(1+3u)du=
v=1+3u, u=(v-1)/3, dv=3dx
=1/6∫ ((v-1)/3)/vdv
=1/6∫ (v-1)/(3v)dv
=1/6∫ (1/3-1/(3v))dv
=v/18-ln(v)/3+C
然后把x带进去就显然了.
∫ 1/(1+3sqrt(x))dx=
x=u^2, du=2udx
=1/2∫ u/(1+3u)du=
v=1+3u, u=(v-1)/3, dv=3dx
=1/6∫ ((v-1)/3)/vdv
=1/6∫ (v-1)/(3v)dv
=1/6∫ (1/3-1/(3v))dv
=v/18-ln(v)/3+C
然后把x带进去就显然了.
展开全部
(1)
z=e^(xy). cos(x+y)
dz = [-sin(x+y) .(dx +dy ) + cos(x+y). (xdy+ydx) ] .e^(xy)
(2)
let
u=√x
2udu = dx
∫dx/(1+3√x)
=∫ [2u/(1+3u)] du
=∫ [ (2/3) (1+3u) - 2/3]/(1+3u) du
=∫ { 2/3 - (2/3) [1/(1+3u)] } du
=(2/3)u - (2/9)ln|1+3u| + C
=(2/3)√x - (2/9)ln|1+3√x| + C
z=e^(xy). cos(x+y)
dz = [-sin(x+y) .(dx +dy ) + cos(x+y). (xdy+ydx) ] .e^(xy)
(2)
let
u=√x
2udu = dx
∫dx/(1+3√x)
=∫ [2u/(1+3u)] du
=∫ [ (2/3) (1+3u) - 2/3]/(1+3u) du
=∫ { 2/3 - (2/3) [1/(1+3u)] } du
=(2/3)u - (2/9)ln|1+3u| + C
=(2/3)√x - (2/9)ln|1+3√x| + C
追问
感激不尽!
本回答被提问者采纳
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