法1因为不定积分∫(x+sinx)/(1+cosx)dx=∫[x+2sin(x/2)cos(x/2)]/[2cos²(x/2)]dx=∫[x/(2cos²(x/2))]dx+∫[2sin(x/2)cos(x/2)]/[2cos²(x/2)]dx=∫xdtan(x/2)+∫tan(x/2)dx=xtan(x/2)-∫t...