ln[x+√(1+x2)] 的导数怎么解?

(ln[x+√(1+x²)])'
=[x+√(1+x²)]'/[x+√(1+x²)]
=[1+1/2*2x/√(1+x²)]/[x+√(1+x²)]
=[1+x/√(1+x²)]/[x+√(1+x²)]
=[x+√(1+x²)]/{[x+√(1+x²)]*√﹙1＋x²）}
=1/√﹙1＋x²）