Question

该积分最后一步的ln是如何化简的？

我得到的是ln(3+2√2)，答案是怎么把1/2弄进去的

Answer 1

[π/(2√2)] [ ln|(√2+t)/(√2-t)|] |(0->1)
=[π/(2√2)] . ln[(√2+1)/(√2-1)]
=[π/(2√2)] . ln[(√2+1).(√2+1)]
=[π/(2√2)] . ln[(√2+1)^2]
=[π/(2√2)] . { 2 ln(√2+1) }
=(π/√2) . ln(√2+1)

Answer 2

Alogb C =logb C^A。1/2可以变为指数的多次方，既1/2ln D=ln D^1/2