dx J01+√ 解 x=1,x=12,dx=2tdt,x=0,t=0x=1,1=1 -12 -2u-1+M =2-4nl:0日 =2ln2-1 1212 例 5 求 1 0 1 x dx + x  解 令 2 x t x t dx tdt x t x t = = = = = = = , , 2 , 0, 0; 1, 1 1 1 0 0 2 1 1 x t dx tdt x t = + +   2 1 0 1 1 2 1 t tdt t − + = +  1 0 1 2 ( 1 ) 1 t dt t = − + +  2 1 1 2( ln 1 ) 2 0 = − + + t t t = − 2ln 2 1