Question

求计算定积分ln(x+√(x^2+1))dx ,上限1,下限0

Answer 1

∫ln(x+√(1+x^2))dx=xln(x+√(1+x^2))-∫xdln(x+√(1+x^2)

=xln(x+√(1+x^2)-√(1+x^2)+C

∫[0,1]ln(x+√(1+x^2)dx=ln(1+√2)-√2+1