Question

计算不定积分∫x√x2+1dx

Answer 1

∫x√(x^2+1)dx

=(1/2)∫√(x^2+1)dx^2

=(1/2)∫(x^2+1)^(1/2)d(x^2+1)

=(1/2)*[(x^2+1)^(1/2+1)/(1/2+1)]+C

=(1/3)(x^2+1)^(3/2)+C

=(1/3)*(x^2+1)/√(x^2+1)+C