Question

计算：定积分∫（在上2,在下 0）ln(x+√x^2+1） dx 注 跟号包括x^2+1 ..

Answer 1

∫[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