∫1/[2x+√(1-x^2)] dx的值

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∫1/[2x+√(1-x^2)] dx的值
∫1/[2x+√(1-x^2)] dx的值

∫1/[2x+√(1-x^2)] dx的值
令x=sinu,则:u=arcsinx,dx=cosudu.
∴∫{1/[2x+√(1-x^2)]}dx=∫[1/(2sinu+cosu)]cosudu.
引入辅助角t,使cost=2/√5、sint=1/√5.则:t=arcsin(1/√5).
∴∫{1/[2x+√(1-x^2)]}dx
=∫[1/(2sinu+cosu)]cosudu
=(1/√5)∫[1/(sinucost+cosusint)]cos(u+t-t)d(u+t)
=(1/√5)∫{[cos(u+t)cost+sin(u+t)sint]/sin(u+t)}d(u+t)
=(1/√5)cost∫[cos(u+t)/sin(u+t)]d(u+t)+(1/√5)sint∫d(u+t)
=(1/√5)×(2/√5)∫[1/sin(u+t)]d[sin(u+t)]+(1/√5)×(1/√5)(u+t)
=(2/5)ln|sin(u+t)|+(1/5)[arcsinx+arcsin(1/√5)]+C
=(2/5)ln|sinucost+cosusint|+(1/5)arcsinx+C
=(2/5)ln|(2/√5)x+(1/√5)√(1-x^2)|+(1/5)arcsinx+C
=(2/5)ln|2x+√(1-x^2)|-ln√5+(1/5)arcsinx+C
=(2/5)ln|2x+√(1-x^2)|+(1/5)arcsinx+C