Question

求最后一步的详细计算过程，就是怎么算出√2ln√2 +1的

Answer 1

见下图：

Answer 2

f(x) = (1/√2)ln| [√(1-x) -√2]/[√(1-x) +√2]|
f(1)
= (1/√2)ln| [√(1-1) -√2]/[√(1-1) +√2]|
=(1/√2)ln| -√2/√2 |
=(1/√2)ln1
=0
f(0)
= (1/√2)ln| [√(1-0) -√2]/[√(1-0) +√2]|
= (1/√2)ln| (1 -√2)/(1 +√2) |
=(1/√2)ln| (1 -√2)(√2-1) |
=(1/√2)ln(3-2√2)
(1/√2)ln| [√(1-x) -√2]/[√(1-x) +√2]| |(0->1)
=f(1) -f(0)
=-(1/√2)ln(3-2√2)
=-(√2/2)ln(3-2√2)
=√2ln [1/√(3-2√2)]
=√2ln [1/(√2-1)]
=√2ln [√2+1)]
//
3-2√2 = (√2-1)^2
√(3-2√2) =√2-1