∫1/(9+4x^2) dx
令x=3/2*tanz,dz=3/2*sec^2z dz
原式=3/2*∫sec^2z/(9+4*9/4*tan^2z)
=3/2*∫sec^2z/(9sec^2z) dz
=1/6*z+C
=(1/6)arctan(2x/3)+C
求积分 ∫（1/9+(4x的平方）dx
∫1/(9+4x^2) dx
令x=3/2*tanz,dz=3/2*sec^2z dz
原式=3/2*∫sec^2z/(9+4*9/4*tan^2z)
=3/2*∫sec^2z/(9sec^2z) dz
=1/6*z+C
=(1/6)arctan(2x/3)+C