Question

求解微积分

求解微积分5.6小问

Answer 1

（5）
∫(0->a) dx/[(a^2+x^2).√(a^2+x^2)]
let
x=atanu
dx=a(secu)^2. du
x=0, u=0
x=a, u=π/4
∫(0->a) dx/[(a^2+x^2).√(a^2+x^2)]
=∫(0->π/4) a(secu)^2. du /[(a^3 (secu)^3]
=(1/a^2)∫(0->π/4) cosu du
=(1/a^2)[sinu]|(0->π/4)
=√2/(2a^2)
(6)
∫(√e->e) dx/{x√[lnx(2-lnx)]}
let
x=e^t
dx=e^t dt
x=√e , t=1/2
x=e, t=1
∫(√e->e) dx/{x√[lnx(2-lnx)]}
=∫(1/2->1) e^t dt/{e^t .√[t(2-t)] }
=∫(1/2->1) dt/√[t(2-t)]
=∫(-π/6->0) du
=[u]|(-π/6->0)
=π/6
consider
t(2-t)= 2t -t^2 = 1-(t-1)^2
let
t-1 = sinu
dt = cosu du
t=1/2 , u = -π/6
t=1, u=0