(6)∫(tan x)sec2xdr=∫f(tanx)dtanx (7)∫fe*)e*dr=∫fe*)de ⑧jfnx)'dx=jfna)dnx s求∫dk 解原式=2e3dx引e2d3d=  (6) f (tan x)sec xdx 2 dtan x =  f e e x x x (7) ( ) d x de =  x x f x d 1 (8) (ln ) dln x 例8. 求 d . 3 x x e x  解: 原式 = e x x 2 d 3  d(3 ) 3 2 3 e x x  = e C x = + 3 3 2