⑩串紫学 Teaching Plan on Advanced Mathematics 例3 +Bx+C (1+2x)(+x2)1+2x1+x2 解1=A(1+x2)+(Bx+C(1+2x 整理得1=(4+2B)x2+(B+2C)x+C+A, A+2B=0, B+2C=0,→小4 B C 5 5 A+C=1, 21 55 (1+2x)(1+x2)1+2x1+x2
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例3 . 1 5 1 5 2 1 2 5 4 2 x x x + − + + + = (1 2 )(1 ) 1 2 + x + x 1 (1 ) ( )(1 2 ), 2 = A + x + Bx + C + x 1 ( 2 ) ( 2 ) , 2 = A+ B x + B + C x + C + A + = + = + = 1, 2 0, 2 0, A C B C A B , 5 1 , 5 2 , 5 4 A = B = − C = , 1 2 1 2 x Bx C x A + + + + = (1 2 )(1 ) 1 2 + x + x 整理得 解
⑩串紫学 Teaching Plan on Advanced Mathematics 例4求积分 .dx xlr 解 dx xlx xX d x d x d x =Ix-—-ln(x-1)+C
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例4 求积分 . ( 1) 1 2 dx x x − dx x x − 2 ( 1) 1 dx x x x − − − = + 1 1 ( 1) 1 1 2 dx x dx x dx x − − − = + 1 1 ( 1) 1 1 2 ln( 1) . 1 1 ln x C x x − − + − = − 解
⑩串紫学 Teaching Plan on Advanced Mathematics 例5求积分 (1+2x)(1+x x十 解 5 (1+2x)+x2)1+2x d x t 1+y 2 ln(1+2x) 1【2xdx+ d x 51+x 2 2 51+x -In(1+2x)--In(1+x)+arctan+C. 5 5
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例5 求积分 解 . (1 2 )(1 ) 1 2 + + dx x x dx x x dx x + − + + + = 2 1 5 1 5 2 1 2 5 4 + + dx (1 2x)(1 x ) 1 2 dx x dx x x x + + + = + − 2 2 1 1 5 1 1 2 5 1 ln(1 2 ) 5 2 arctan . 5 1 ln(1 ) 5 1 ln(1 2 ) 5 2 2 = + x − + x + x + C
⑩串紫学 Teaching Plan on Advanced Mathematics 例6求积分 1+e2+e3+e 解令t=e6→x=6lnt,dx=-lt, dt 1+t3+t2+t 1+e2+e3+e 633t+3 +D)(1+ t1+t1+t2
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例6 求积分 解 . 1 1 2 3 6 dx e e e x x x + + + 令 6 x t = e x = 6lnt, , 6 dt t dx = dx e e e x x x + + + 2 3 6 1 1 dt t t t t 6 1 1 3 2 + + + = dt t t t + + = (1 )(1 ) 1 6 2 dt t t t t + + − + = − 2 1 3 3 1 6 3