解答:解 令t=arctanx,即x=tant,则 ∫ xearctanx (1+x2)3 2 dx=∫ tant?et (1+tan2t)3/2 ?sec2tdt=∫etsintdt 而∫etsintdt=∫etd(-cost)=-costet+∫etcostdt=-costet+∫etdsint=-costet+sintet-∫etsintdt ∴∫etsintdt=1 2 et(sint?cost)+C.(其中C为任意常数) ∫ xearctanx (1+x2)3 2 dx=1 2 et(sint?cost)+C=1 2 earctanx( x 1+x2 ?1 1+x2 )+C