ln（x+√x^2+1）的积分
问题描述：
ln（x+√x^2+1）的积分
答
∫[0,2] ln[x+√(x^2+1)] dx
=xln[x+√(x^2+1)][0,2]-∫[0,2] xdln[x+√(x^2+1)]
=2ln(2+√5)-∫[0,2] x/[x+√(x^2+1)]*[x+√(x^2+1)]'dx
=2ln(2+√5)-∫[0,2] x/[x+√(x^2+1)]*[1+x/√(x^2+1)]dx
=2ln(2+√5)-∫[0,2] x/√(x^2+1)dx
=2ln(2+√5)-√(x^2+1)[0,2]
=2ln(2+√5)-√5+1