定理1设∫f(n)dm=F(u)+c,则有 Sflp(x)ldo(x)=F((x)+C 运用凑微分法其难点在于原题并未指明应该把哪 部分凑成dq(x),这需要解题经验,记住下列一些 微分式,会给我们解题带来很大帮助 dx dx=d(ax+b), xdx==d(x ), =2d(√x) 2 ed=d(e), sin xx =-d(cos x),dx=d(n x) Economic-mathematics 48-6 Wednesday, February 24, 2021
Economic-mathematics 48- 6 Wednesday, February 24, 2021 运用凑微分法其难点在于原题并未指明应该把哪 一部分凑成 d(x),这需要解题经验,记住下列一些 微分式 , 会给我们解题带来很大帮助． ( )， 1 d ax b a dx = + ( )， 2 1 2 xdx = d x 2d( x)， x dx = e x dx = d(e x )， (ln | |)， 1 dx d x x sin xdx = −d(cos x)， = 定理1 设 f (u)du = F(u) + c，则有 f[(x)]d(x) = F((x)) + C
例1求 3+2 11 解 (3+2x), 3+2x23+2x 3+2x23+2t(3+2)h令3+2x= Inu+C=In(3+2x)+C 2 般地∫f(ax+b)c=订n(a+b0(ax+b) Economic-mathematics 48-7 Wednesday, February 24, 2021
Economic-mathematics 48- 7 Wednesday, February 24, 2021 例1 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 = [ ( + ) ( + ) 1 f ax b d ax b a 一般地 令 3 + 2x = u
例2求∫ rnx 解 d(ln x) xnx Inx du u=Inx =Inu+c In Inx+c Economic-mathematics 48-8 Wednesday, February 24, 2021
Economic-mathematics 48- 8 Wednesday, February 24, 2021 例2 求 . ln 1 dx x x 解 dx x x ln 1 (ln ) ln 1 d x x = = du u = ln x u 1 = lnu +C = lnln x +C
例3求 a x 解 dx -+x 1+ 2 ∫1d(5)=1 arctan+C 41+()2 Economic-mathematics 48-9 Wednesday, February 24, 2021
Economic-mathematics 48- 9 Wednesday, February 24, 2021 例3 求 . 1 2 2 dx a x + 解 dx a x + 2 2 1 dx a a x + = 2 2 2 1 1 1 ( ) 1 ( ) 1 1 2 a x d a a x + = arctan . 1 C a x a = +
例4求∫i2xb 解1∫ sin rdt I jsin 2xd(2x)=cos 2x+C 解2∫sim2xd=2」 sin x cosxd 2 sin xd(sin x)=(sin x)+C 解3∫sim2xdx=2s in x cos xdx -2 cos xd(cos x)=-(cosx)+C 可见,不同方式的积分可能会有不同形式的结果 Economic-mathematics 48-10 Wednesday, February 24, 2021
Economic-mathematics 48- 10 Wednesday, February 24, 2021 例4 求 sin 2 . xdx 解 1 sin 2xdx= sin 2 (2 ) 2 1 xd x cos2 ; 2 1 = − x + C 解 2 sin 2xdx = 2 sin xcos xdx = 2 sin xd(sin x) (sin ) ; 2 = x + C 解 3 sin 2xdx = 2 sin xcos xdx = − 2 cos xd(cos x) (cos ) . 2 = − x +C 可见，不同方式的积分可能会有不同形式的结果