Integration problem

Let t = 5 - 4e^(3x)
dt = -12e^(3x) dx

When x = 0, t = 1
When x = 1, t = 5 - 4e^3

Therefore,
1
S [ -12e^(3x) / ( (5 - 4e^(3x)) ^(4/3) ) ] dx
0

5 - 4e^3
= S [ dt / t^(4/3) ]
1


= [ -3t(^-1/3) ] upper = 5 - 4e(^3), lower = 1

= 3 - 3 (5 - 4e(^3) ) (^-1/3)

let t=5-4e^(3x)
dt=-12e^(3x)dx
So
1
S((-12e^(3x))/((5-4e^(3x))^(4/3))) dx
0
1
=S 1/t^(4/3) dt
0
1
=S t^(-4/3) dt
0
=-3t^(-1/3) |1,0
但0^(-1/3) 是無限大
不知你是否打錯了問題?