请问利用求第一类换元法求不定积分∫dx/(sinx+cosx)怎么做?

设t=tan(x/2),则x=2arctant,dx=2/(1+t²) dt∫dx/(sinx+cosx)=∫1/[2t/(1+t²)+(1-t²)/(1+t²)]·2/(1+t²) dt=∫2/(-t²+2t+1) dt=-2∫1/(t²-2t-1) dt=-2∫1/[(t-1-√2)(t-1+√2)] d...