例16、设f(x)=J1 x In t dtx>0,求f(x)+ 解-::于 dt Inx 1+ nX Inx Inx 1+xx(x+1) Inx f(x)+ n2x+c令x=1c= (∵f(1)=0) f(x)+|=h2x 例17、设(x)=0dt求∫。(x)d 解: fxdx=∫6xd(x-x)=(x-x)fx)-∫。(x-mdfx) SInX sinxdx= 212 例 16、设 ( ) dt x 0 1 t ln t f x x 1 + = ,求 ( ) + x 1 f x f 解: ( ) + + + = + x 1 1 x 1 dt 1 t lnt dt 1 t ln t x 1 f x f − + + + = 2 x 1 x 1 1 x 1 ln 1 x lnx ( ) x lnx x x 1 lnx 1 x lnx = + + + = ( ) ln x c 2 1 dx x lnx x 1 f x f 2 = = + + 令 x =1 c = 0 (∵ f(1)= 0 ) ∴ ( ) ln x 2 1 x 1 f x f 2 = + 例 17、设 − = x 0 dt π t sint f(x) 求 π 0 f(x)dx 解: = − = − − − π 0 π 0 π 0 π 0 f(x)dx f(x)d(x π) (x π )f(x) (x π )df(x) dx sinxdx 2 π - x sinx (x -π ) π 0 π 0 = = =