let
y=-x +6
dy = -dx
x=2, y=4
x=4, y=2
I=∫(2->4) √ln(9-x) /[ √ln(9-x) +√ln(x-3) ] dx
=∫(4->2) √ln(y-3) /[ √ln(9-y) +√ln(y-3) ] (-dy)
=∫(2->4) √ln(x-3) /[ √ln(9-x) +√ln(x-3) ] dx
2I= ∫(2->4) dx
= 2
I=1
=>
∫(2->4) √ln(9-x) /[ √ln(9-x) +√ln(x-3) ] dx =1
题目是不是有误?在x∈[2,4]时,ln(x-3)及√[ln(x-3)]不一定有意义哦。