例7.求1=∫ earctanx dx.先换元后分部 解：令t=arctanx,即x=tant,则 1-,ca-j小e'sd, =e'sint-∫e'sintdt V1+ =e'sint+e'cost-Je'cosidt 故I=(sint+cost）)e'+C arctan x+C 例7. 求 d . I x  = 2 3 (1 ) 2 + x 先换元后分部 令 t = arctan x, 即 x = tant, 则  = t e I t 3 sec sec t d t 2  e t t t cos d  = = e t − t sin e t t t sin d  e t t = sin e t t t cos d  e t − t + cos 故 I t t e C t = (sin + cos ) + 2 1   = 2 1 x e arctan t x 1 2 1+ x 2 1 x x + 2 1 1 + x + e C x +   arctan 解：