Question

高数第二题

Answer 1

令arctanx=t,则x=tant,dx=sec²tdt

∫xe^arctanx/(1+x^2)^3/2dx

=∫tante^t/(1+tan^2;t)^3/2*sec²tdt

=∫tante^t/sec ³t sec ²tdt

=∫tante^t/sectdt

=∫sinte^tdt (1)

=-∫e^tdcost

=-coste^t+∫coste^tdt

=-coste^t+sinte^t-∫sinte^tdt (2)

由 (1)(2)得

∫sinte^tdt =1/2( sinte^t-coste^t ) +C

=1/2( sint-cost)e^t +C

=1/2cost(tant-1)e^t +C

=1/2*1/√(tan²t+1)*(tant-1)e^t +C

=1/2*1/√(x²+1)*(x-1)e^arctanx+C

=√(x²+1)*(x-1)e^arctanx/(x²+1)+C