()Pn(x)edx型 例1求xedx ∫xeak=∫x(e*)y'dk=xe*-∫e'd 解 =xe*-ex+c 例2.求「(x2+1)ed 解∫(x2+1)e'd=∫(x2+1(e)'d =(x2+1)e*-∫e*2xd =(x2+l)e"-2Jxde" =(x2+1)e-2(xe-∫ed) =(x2+1)ex-2xe*+2e'+c =(x2-2x+3)e*+c 13 13 ㈠ ( ) ax  P x e dx n 型 例 1 求 x xe dx  解 ( )' x x x x xe dx x e dx xe e dx = = −    x x = − + xe e c 例 2.求 2 ( 1) x x e dx +  解 2 2 ( 1) ( 1)( )' x x x e dx x e dx + = +   2 ( 1) 2 x x = + − x e e xdx  2 ( 1) 2 x x = + − x e xde  2 ( 1) 2( ) x x x = + − − x e xe e dx  2 ( 1) 2 2 x x x = + − + + x e xe e c 2 ( 2 3) x = − + + x x e c