令y = ln[x + √(x² + 1)],确保y是奇函数才存在反函数y⁻¹e^y = x + √(x² + 1)√(x² + 1) = e^y - xx² + 1 = e^(2y) - 2xe^y + x²2xe^y = e^(2y) - 1x = [e^(2y) - 1]/(2e^y)所...