第1个回答 2019-03-06
如图所示
第2个回答 2019-03-06
∫(0->2) dx/[(x+1)^(1/2) +(x+1)^(3/2)]
let
u = (x+1)^(1/2)
du =(1/2)(x+1)^(-1/2) dx
dx = 2u du
x=0, u=1
x=2, u=√3
∫(0->2) dx/[(x+1)^(1/2) +(x+1)^(3/2)]
=∫(1->√3) 2u du/( u +u^3)
=∫(1->√3) 2 du/( 1 +u^2)
=2[arctanu]|(1->√3)
=2( π/3 -π/4)
=π/6追问
let
u = (x+1)^(1/2)
du =(1/2)(x+1)^(-1/2) dx
dx = 2u du
x=0, u=1
x=2, u=√3
∫(0->2) dx/[(x+1)^(1/2) +(x+1)^(3/2)]
=∫(1->√3) 2u du/( u +u^3)
=∫(1->√3) 2 du/( 1 +u^2)
=2[arctanu]|(1->√3)
=2( π/3 -π/4)
=π/6追问
x=0, u=1
x=2, u=√3
换元后 要把上下限带进去换个新的是吧
换元后 要把上下限带进去换个新的是吧：是
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