例4求 (1+x) 解:原式 y+1-1 (1+x)3 (1+x)3(+砂少h x+1 k(1+x) 1+x) (1+x) 1+x2(1+以+C 十例4 求 dx x x  + 3 (1 ) 解: 原式= dx x x  + + − 3 (1 ) 1 1 dx x x x  + − + + = ] (1 ) 1 (1 ) 1 [ 3 3 ] (1 ) (1 ) 1 (1 ) 1 [ 2 3 d x x x + + − + =  C x x + + + + = − 2 2(1 ) 1 1 1