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.
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第1个回答 2019-06-28
(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追问
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！
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