2道简单 求微积分题

微分

d{e^[(2x+1)^(1/2)]}

= e^[(2x+1)^(1/2)]d[(2x+1)^(1/2)]

= e^[(2x+1)^(1/2)](2x+1)^(-1/2)dx

d{sin(x)e^x}

= sin(x)d(e^x) + e^xd[sin(x)]

= sin(x)e^xdx + e^xcos(x)dx

= [sin(x) + cos(x)]e^xdx

积分
令 u = (2x+1)^(1/2), du = (2x+1)^(-1/2)dx, dx = udu.

Se^[(2x+1)^(1/2)]

= Sue^udu

= ue^u - Se^udu

= ue^u - e^u + C

= (u-1)e^u + C

= [(2x+1)^(1/2) - 1]e^[(2x+1)^(1/2)] + C

记 I = Ssin(x)e^xdx

I = sin(x)e^x - Se^xcos(x)dx

= sin(x)e^x - e^xcos(x) - Se^xsin(x)dx

= [sin(x) - cos(x)]e^x - I,

I = 0.5[sin(x) - cos(x)]e^x + C.

C = const.

1.∫e^√2x+1 dx=∫e^t d(t^2/2-1/2) 设√2x+1=t
=∫te^t dt
=te^t-∫e^t dt 分部积分
=te^t-e^t+c c是常数
=√2x+1e^√2x+1-e^√2x+1+c
2.∫sinx*e^x dx=(-cosx*e^x)+∫cosx*e^x dx 分部积分
∫cosx*e^x dx=sinx*e^x-∫sinx*e^x dx 分部积分
两式相减:解出∫sinx*e^x dx=1/2(-cosx+sinx)e^x+c c是常数

见图

1.∫e^√2x+1 dx=∫e^t d(t^2/2-1/2) 设√2x+1=t
=∫te^t dt
=te^t-∫e^t dt
=te^t-e^t+c
=√2x+1e^√2x+1-e^√2x+1+c
2.∫sinx*e^x dx=(-cosx*e^x)+∫cosx*e^x dx
∫cosx*e^x dx=sinx*e^x-∫sinx*e^x dx
解出∫sinx*e^x dx=1/2(-cosx+sinx)e^x+c

简单

微分1. d(e^√2x+1)=e^√2x+1d(√2x+1)=(e^√2x+1)/(√2x+1)dx
2. d(sinx·e^x)=sinxd(e^x)+e^xd(sinx)=e^x（sinx+cosx)dx
积分1.设(√2x+1)=t,则积分可转化：∫e^td(t^2/2-1/2)=∫te^tdt
运用分部积分法设：dv=e^tdt u=t
∫te^tdt=te^t-∫e^t dt =te^t-e^t+C
反换元得=√2x+1e^√2x+1-e^√2x+1+C
2.运用分部积分法设:u=sinx dv=e^xdx ∫sinxe^xdx=sinxe^x-∫e^xdsinx
=sinxe^x-∫e^xcosxdx
同理运用分部积分法∫e^xcosxdx=e^xcosx-∫sinxe^xdx
代入上式得∫sinxe^xdx=0.5e^x（sinx-cosx)+C