设1+x^2=t dt=2xdx 即xdx=(1/2)dt ∫x√1+x^2 dx =(1/2)∫t^(1/2)dt =(1/2)*(2/3)*t^(3/2) =(1/3)(1+x^2)^(3/2) =(1/3)(1+x^2)√(1+x^2)
