Question

∫[1/√x-x^2]的不定积分怎么求

Answer 1

∫[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