Question

线性代数，图中第五题

Answer 1

设x=tant,t=arctanx,dx=(sect)^2dt
cost=1/√(1+x^2),
sint=x/√(1+x^2)
原式=∫ tant*e^t*(sect)^2dt/([1+(tant)^2]^(3/2)
=∫ tant*e^t*(sect)^2dt/(sect)^3
=∫ sint*e^tdt
=e^t(sint-cost)/2+C
=e^(arctanx)[x/√(1+x^2)-1/√(1+x^2)]/2+C.
=e^(arctanx)/2[(x-1)/√(1+x^2)+C.

对∫ sint*e^tdt用分部积分,
u=sint,v'=e^t,
u'=cost,v=e^t,
∫ sint*e^tdt=e^tsint-∫ e^tcostdt,
∫ e^tcostdt
设u=cost,v'=e^t,
u'=-sint,v=e^t,
∫ e^tcostdt=cost*e^t+∫ e^tsintdt
∴∫ sint*e^tdt=e^tsint-cost*e^t-∫ e^tsintdt,
∴∫ sint*e^tdt=e^t(sint-cost)/2+C.