1.∫[(sec^2x-1/(1+x^2)]dx
=tanx-arctanx+C
2.设x=√3/2tant-1/2
原式=∫(√3/2tant+1/2)/[√3/2*sect*√3/2*sec^2tdt
=∫(√3/2*sint/cos^2t+1/2*1/cost)dt
=√3/2*1/cost+1/4∫[(1/(1-sint)+1/(1+sint)]dsint
=√3/2*1/cost+1/4[ln(1+sint)-ln(1-sint)]+C'
=√(x^2+x+1)+1/4ln{[[1+(x+1/2)/√(x^2+x+1)]-ln[[1-(x+1/2)/√(x^2+x+1)]}+C'
=√(x^2+x+1)+1/2ln[√(x^2+x+1)+x+1/2]+C
(C=-1/4*ln(3/4)+C')