设t=tan(x/2),则x=2arctant,sinx=2t/(1+t²),cosx=(1-t²)/(1+t²),dx=2dt/(1+t²)

　　故∫dx/(1+sinx+cosx)=∫[2dt/(1+t²)]/[1+2t/(1+t²)+(1-t²)/(1+t²)]

　　=∫[2dt/(1+t²)]/[2(1+t)/(1+t²)]

　　=∫dt/(1+t)

　　=ln│1+t│+C(C是积分常数)

　　=ln│1+tan(x/2)│+C.