设√(1+x²)=t x²=t²-1 xdx=tdt
∫(0,1)x^3√(1+x²)dx
=∫(1,√2)(t²-1)t²dt
=(t^5/5-t^3/3）|（1,√2）
=(4√2/5-2√2/3-1/5+1/3)
=(2/15)(√2+1)