Question

高数，用分部积分法求第8题和第11题

Answer 1

8、令t=x^(1/3)，x=t^3，dx=3t^2dt
原式=∫e^t*3t^2dt
=∫3t^2*d(e^t)
=3t^2*e^t-∫e^t*6tdt
=3t^2*e^t-∫6t*d(e^t)
=3t^2*e^t-6t*e^t+∫6e^tdt
=3t^2*e^t-6t*e^t+6e^t+C
=(3t^2-6t+6)e^t+C
=[3x^(2/3)-6x^(1/3)+6]*e^[x^(1/3)]+C，其中C是任意常数
11、令t=lnx，x=e^t，dx=e^tdt
原式=A=∫cost*e^tdt
=∫e^td(sint)
=e^t*sint-∫sint*e^tdt
=e^t*sint+∫e^td(cost)
=e^t*sint+e^t*cost-∫cost*e^tdt
=e^t*(sint+cost)-A
所以A=e^t*(sint+cost)/2+C
原式=A=x*(sinlnx+coslnx)/2+C，其中C是任意常数

Answer 2

(8)令x^(1/3)=t,则x=t³,dx=3t²
原式=3∫t²e^tdt
=3(t²e^t-2∫te^tdt)
=3[t²e^t-2(te^t-∫e^tdt)
=3[t²e^t-2te^t+2e^t+C]
=3e^t(t²-2t+2)+C
=自己把x代回去
(11)设lnx=t,则x=e^t,dx=e^tdt
原式=∫coste^tdt
=sinte^t-∫sine^tdt
=sinte^t-[-coste^t-∫(-cost)e^tdt]
=sinte^t+coste^t-∫coste^tdt
∴2∫coste^tdt=(sint+cost)e^t
∫coste^tdt=(sint+cost)e^t/2+C
=自己代x

Answer 3