求不定积分：$(sin x +cos x)/(1+cos^2x) dx

∫(sin x +cos x)/(1+cos²x) dx=∫(sinx)/(1+cos²(x))dx+∫cosx/(1+cos²x)dx =-∫(dcosx)/(1+cos²x)+∫(dsinx)/(2-sin²x) =-arctan(cosx)+(√2/2)∫(d(sinx/√2))/[1-(sinx/√2)²] 现在令t=sinx/√2 =-arctan(cosx)+(√2/2)∫dt/[1-t²] 此步证明在下方 =-arctan(cosx)+(√2/4)(-ln|sinx/√2-1|+ln|sinx/√2+1|) =-arctan(cosx)+(√2/4)ln[|(sinx/√2+1)/(sinx/√2-1)|] ∫dt/[1-t²] =(1/2)*∫[1/(1-t)+1/(1+t)]dt =(1/2)*{-∫1/(t-1)d(t-1)+∫1/(1+t)d(1+t)} =(1/2)*(-ln|t-1|+ln|t+1|)