答：设x=sint原式=∫ [1/(1+cost)]d(sint)=∫ [cost/(1+cost)]dt=∫ dt -∫ 1/(1+cost) dt=t -∫ 1/[cos(t/2)]^2 d(t/2)=t - tan(t/2)+C=t- 2sin(t/2)cos(t/2)/2[cos(t/2)]^2+C=t-sint/(1+cost)+C=arcsinx-x/[1+√(1...