原题是∫[0→ln2] e^(2x) * √[e^(2x) - 1] dx吧?
令y = e^(2x)、2x = lny、x = (1/2)lny、dx = (1/2)(1/y) dy
当x = 0、y = 1
当x = ln2、y = 4
原式 = ∫[1→4] y√(y - 1) * [(1/2)(1/y) dy]
= (1/2)∫[1→4] √(y - 1) dy
= (1/2)∫[1→4] √(y - 1) d(y - 1)
= (1/2)(2/3)(y - 1)^(3/2) |[1→4]
= (1/3)(4 - 1)^(3/2) - 0
= 3^(- 1) * 3^(3/2)
= 3^(3/2 - 1)
= √3