∫1／(x+√1-x²)dx
做三角代换：令x=sint,则√1-x²=cost,dx=costdt
原式=∫cost／(sint+cost)dt=(1/2)∫ (cost+sint+cost-sint)／(sint+cost)dt
=(1/2)∫ (cost+sint)／(sint+cost)dt+(1/2)∫ (cost-sint)／(sint+cost)dt
=(1/2)∫ 1dt+(1/2)∫ 1／(sint+cost)d(sint+cost)
=(1/2)t+(1/2)ln|sint+cost|+C
=(1/2)arcsinx+(1/2)ln|x+√1-x²|+C