Question

请问第六题怎么做？

Answer 1

let
u=√x
2u du = dx
x=0, u=0
x=3, u=√3
∫(0->3) arcsin√[x/(1+x)] dx
=[x.arcsin√[x/(1+x)] ]|(0->3) -∫(0->3) x darcsin√[x/(1+x)]
=π - ∫(0->3) x.[1/√{ 1 - [x/(1+x)] } ] . { 1/(2√[x/(1+x)] ) } . [ 1/(1+x)^2 ] dx
=π - (1/2)∫(0->3) √x /(1+x) dx
=π - (1/2)∫(0->√3) [u /(1+u^2)] (2udu)
=π - ∫(0->√3) u^2 /(1+u^2) du
=π - ∫(0->√3) [ 1- 1/(1+u^2)] du
=π - [ u- arctanu]|(0->√3)
=π - ( √3- π/3 )
=4π/3 - √3

Answer 2

令u=arcsin√...
则sin²u=x/(1+x)=1－1/(1+x)
x=1/(1－sin²u)－1=tan²u
=∫(0.π/3)udtan²u
=utan²u－∫tan²udu
=utan²u－(tanu－u)
=π－(√3－π/3)
=4π/3－√3