y=∫(0.x) 【（3t+1）/（t^2－t+1）】dt在[0,1]上的最大值
问题描述：
y=∫(0.x) 【（3t+1）/（t^2－t+1）】dt在[0,1]上的最大值
(0.x)为积分下限和积分上限
最大值为5pai/3根号3
答
y=∫(0->x) (3t+1)/(t^2－t+1）dt
y' =(3x+1)/(x^2－x+1)
=(3x+1)/[(x- 1/2)^2 +3/4] >0
max y = y(1)
max y
=∫(0->1) (3t+1)/(t^2－t+1）dt
= (3/2)∫(0->1) (2t-1)/(t^2－t+1）dt +(5/2)∫(0->1) dt/(t^2－t+1）
= (3/2)ln|t^2－t+1| |(0->1) +(5/2)∫(0->1) dt/(t^2－t+1）
=(5/2)∫(0->1) dt/(t^2－t+1)
consider
t^2－t+1 = (t-1/2)^2+3/4
let
t-1/2 = (√3/2)tany
dt =(√3/2)(secy)^2 dy
t=0 ,y = -π/6
t=1,y=π/6
∫(0->1) dt/(t^2－t+1)
=∫(-π/6->π/6) dt/(t^2－t+1)
=(2√3/3) ∫(-π/6->π/6) dy
=2√3π/9
max y
=(5/2)∫(0->1) dt/(t^2－t+1)
=5√3π/9