应该是1/2log以2为底[2x+2√(x^2-1)]+log以2为底[√(x+1)-√(x-1)]
1/2log以2为底[2x+2√(x^2-1)]+log以2为底[√(x+1)-√(x-1)]=log以2为底根号[2x+2√(x^2-1)]+log以2为底[√(x+1)-√(x-1)]=log以2为底根号[√(x+1)+√(x-1)]+log以2为底[√(x+1)-√(x-1)]
=log以2为底[√(x+1)+√(x-1)]*[√(x+1)-√(x-1)]=log以2[(x+1)-(x-1)]=log以2底(2)=1