令t=tanx
原式=∫1/[(1+t)(1+t^2)]dt
=(1/2)∫1/(1+t)dt-(1/2)∫(t-1)/(1+t^2)dt
=(1/2)ln|1+t|-(1/2)∫(t-1)/(t^2+1)dt
…………①
而∫(t-1)/(t^2+1)dt
=(1/2)∫1/(t^2+1)d(t^2+1)-∫1/(t^2+1)dt
=(1/2)ln|t^2+1|-arctant+C………②
将②式代回①式即可.