例2求 3+2x 解 3+2x23+2x (3+2x), 3+2x23+2x (3+2x)x ==?lm+C=lm(3+2x)+C. 2 般地∫f(ax+b)x=∫/(a)ohl u=ax+b
例2 求 . 3 2 1 dx x + 解 (3 2 ) , 3 2 1 2 1 3 2 1 + + = + x x x dx x 3 + 2 1 x dx x (3 2 ) 3 2 1 2 1 + + = du u = 1 2 1 = lnu + C 2 1 ln(3 2 ) . 2 1 = + x + C f (ax + b)dx = u du u=ax+b f a [ ( ) ] 1 一般地
常用的几种配元形式 (1)f(ax+b)dx=f(ax+b)d(ax+b) (2)|(x"ar≈1 f(x")dx n (3)[f(x2) dx三 f(x") dx x 万能凑幂法 4)f(sin x ) cos xdx= f(sin x)dsin x (5)If(cos x)sin xdx=- f(cos x)dcosx
常用的几种配元形式: + = (1) f (ax b)dx d(ax + b) a 1 = − f x x x n n (2) ( ) d 1 n dx n 1 = x x f x n d 1 (3) ( ) n dx n 1 n x 1 万 能 凑 幂 法 = (4) f (sin x)cos xdx dsin x = (5) f (cos x)sin xdx − dcos x
(6)f(tanx)sec xdx=I f(tan x)dtanx (7)If(e )e dx= f(e")de )∫mx)dx=J/(mlnx dx 例3求」 r(1+2Inx) 解:原式= dInx 1 rd(1+2Inx' 1+2Inx 2J1+2Inx In 1+2Inx+C
= (6) f (tan x)sec xdx 2 dtan x = f e e x x x (7) ( ) d x de = x x f x d 1 (8) (ln ) dln x 例3. 求 1+ 2ln x dln x 解: 原式 = + = 2 1 2ln x 1 d(1+ 2ln x)
例4求 (1+x) 解 (1+)2fx+1-1 1+x) (1+x)2(1+1)21(1+x) +C, +c x 2(1+x)2 +c 1+x2(1+x
例4 求 . (1 ) 3 dx x x + 解 dx x x + 3 (1 ) dx x x + + − = 3 (1 ) 1 1 ] (1 ) (1 ) 1 (1 ) 1 [ 2 3 d x x x + + − + = 1 2 2 2(1 ) 1 1 1 C x C x + + + + + = − . 2(1 ) 1 1 1 2 C x x + + + + = −
例5求「,,b 2 a +x 解 -+x x)1 arctan -+C 1+
例5 求 . 1 2 2 dx a x + 解 dx a x + 2 2 1 dx a a x + = 2 2 2 1 1 1 + = a x d a a x 2 1 1 1 arctan . 1 C a x a = +