求∫1/[x2√(4+x2)]dx
设x=2tan t，则dx=2dt/cos2t，故 ∫={1/[x2√(4+x2)]}dx =∫1/(4tan2t)•[1/(2/cost)]•(2/cos2t)dt =∫(cost/4sin2t)dt=-(1/4sint)+C =-√(4+x2)/(4x)+C(这里已设x﹥0)．
求∫1/[x2√(4+x2)]dx
设x=2tan t，则dx=2dt/cos2t，故 ∫={1/[x2√(4+x2)]}dx =∫1/(4tan2t)•[1/(2/cost)]•(2/cos2t)dt =∫(cost/4sin2t)dt=-(1/4sint)+C =-√(4+x2)/(4x)+C(这里已设x﹥0)．
