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