1例3求dxx(x-1)1ABC解-2x(x-1)(x-1)xx-1赋值(1)1= A(x -1)2 + Bx +Cx(x-1)代入特殊值来确定系数A,B,C取x=0, =A=1取x=1,=B=1取 x = 2, 并将 A,B值代入 (I) →C=-11111x(x -1)2(x -1)2x-1x
x x x d ( 1) 1 2 − = + x A 1 ( 1) ( 1) 2 = A x − + Bx +Cx x − 代入特殊值来确定系数 A,B,C 取 x = 0, A = 1 取 x = 1, B = 1 取 x = 2, 并将 A,B 值代入 C = −1 1 1 ( 1) 1 1 2 − − − = + x x x 2 ( 1) 1 − x x 例3 求 2 ( 1) 1 x x − 解 (1) (1) + − 2 (x 1) B x − 1 C 赋值
1dxx(x -1)11x(x -1)xx1于是IxdxxIn|x-1 + C=In|xx-1
于是 x x x d ( 1) 1 2 − − − − = + x x x x x x d 1 1 d ( 1) 1 d 1 2 x C x x − − + − = − ln 1 1 1 ln | | x x x x d 1 1 ( 1) 1 1 2 − − − = + x x x d ( 1) 1 2 − 1 1 ( 1) 1 1 2 − − − = + x x x 2 ( 1) 1 x x −
1例4 求dx(1 + 2x)(1 + x1ABx + C解1+x?(1+ 2x)(1+ x21+2x二次质因式1 = A(1 + x2)+(Bx +C)(1+ 2x)1=(A+2B)x2 +(B+2C)x+C+ A比较系数A+2B=0,2B+2C=0. =75A+C=1
1 (1 ) ( )(1 2 ) 2 = A + x + Bx + C + x 1 = (A+ 2B)x + (B + 2C)x + C + A 2 + = + = + = 1, 2 0, 2 0, A C B C A B 5 1 , 5 2 , 5 4 A = B = − C = 例4 求 d . (1 2 )(1 ) 1 2 + + x x x 解 = (1+ 2 )(1+ ) 1 2 x x + + x A 1 2 2 1+ x Bx +C 比较系数 二次质因式 比较系数
214x+15551+x?(1+ 2x)(1 + x2)1+2x42x+155dx:dx +dx(1 + 2x)(1 + x2.2+2x1+X22xIn(+ 2xdx +dx55:1+x21In /1+2xl-ln l= arctan x + C555
x x x x x d 1 5 1 5 2 d 1 2 5 4 2 + − + + + = + + x x x d (1 2 )(1 ) 1 2 = ln(1+ 2 ) − 5 2 x ln |1 2 | 5 2 = + x 2 1 5 1 5 2 1 2 5 4 x x x + − + + + = (1 2 )(1 ) 1 2 + x + x ln |1 | 5 1 2 − + x + arctan x +C 5 1 + + x x x d 1 2 5 1 2 x x d 1 1 5 1 2 +