∫ x²√a²+x²dx

分部积分 ∫ x²√a²+x²dx =1/3x(a²+x²)^(3/2)-∫√a²+x²dx 对∫√a²+x²dx,令x=tant,可化为 ∫sec³tdt =∫sectdtant =secttant-∫tantdsect =secttant-∫secttan²tdt =secttant-∫sect(sec²t-1)dt =secttant-∫sec³tdx+∫sectdt =secttant-∫sec³tdt+ln|sect+tant| 2∫sec³tdt=secttant+ln|sect+tant| ∫sec³tdx=(secttant+ln|sect+tant|)/2+C 转换回去：∫√a²+x²dx=(x√a²+x²+ln(x+√a²+x²))/2+C 所以,∫ x²√a²+x²dx= 1/3x(a²+x²)^(3/2)-(x√a²+x²+ln(x+√a²+x²))/2+C A={1,2,3},B={0,1,2} AB=0*1+2*1+3*2=8