由定义,y=sinhx=[e^x-e^(-x)]/2同乘2e^x,得2ye^x=e^2x -1即,e^2x - 2ye^x - 1=0故,e^x=y +/- √(1+y^2)又e^x>0,e^x=y+√(1+y^2),x=ln[y+√(1+y^2)].则arcsinhy=x=ln[y+√(1+y^2)]即arcsinhx=ln[x+√(1+x^2)]...