∫sec2xtanxdx的不定积分结果是

原式=∫ (1/(2^cos^x-1))*sinx/cosx dx
= - ∫ (1/(2^cos^x-1)) / cosx d(cosx)
设u=cosx,化简得
原式=- ∫ 1/((2u^2-1)*u)du
=-∫ 根号2/2(1/(根号2u-1)+1/(1+根号2u))+1/udu
=-1/2(ln(根号2u-1)+ln(根号2u+1))+lnu+C
=ln|cosx|-1/2*ln|cos2x|+C