∫[1/√x-x^2]dx=∫[1/√x√(1-x)]dx=∫[1/√x√(1-(√x)^2]d(√x)^2 (相当于作了代换√x=t=∫[1/√x√(1-(√x)^2)](2√x)d√x=∫[2/√(1-(√x)^2)]d√x=2arcsin√x+C
