Question

ln(x+根号下（x^2+1））怎么求它的反函数啊

Answer 1

令y = ln[x + √(x² + 1)]，确保y是奇函数才存在反函数y⁻¹

e^y = x + √(x² + 1)
√(x² + 1) = e^y - x
x² + 1 = e^(2y) - 2xe^y + x²
2xe^y = e^(2y) - 1
x = [e^(2y) - 1]/(2e^y)
所以反函数y⁻¹为
y = [e^(2x) - 1]/(2e^x)
y = (1/2)[e^(2x) - 1]e^(- x)

Answer 2

设y=in[x+√x^2+1]
e^y=x+√x^2+1
[e^y-x]=√x^2+1
平方得到
[e^y-x]^2=x^2+1
e^2y-2xe^y-x^2-1=0
e^2y-[x^2+2xe^y]-1=0
e^2y-[[x+e^y]^2-e^2y]-1
e^2y-[x+e^y]^2+e^2y-1=0
2e^2y-1=[x+e^y]^2
x=√2e^2y-1]-e^y
xy互换
y=√2e^2x-1]-e^x

Answer 3

反函数就是（x^2+1））的平方