∫ 1/√[（4x^2+9)^3] dx
解设x=3/2*tant,则原式化为
∫ 1/√[（9tant^2+9)^3] d3/2tant
=3/2∫ 1/√[（9(tant)^2+9)^3]*1/(cost)^2 dt
=1/18*∫ |cost| dt
=1/18*√4x^2/(9+4x^2)+c
∫√[1-x/x] dx,令√[1-x/x]=t,则x=1/(1+t^2)
=∫t*(-2t)/(1+t^2)^2 dt
=-∫ 2t^2/(1+t^2)^2 dt
=-∫ 2/(1+t^2)- 2/(1+t^2)^2 dt
=arctant+t/(1+t^2)+c
=arctan√[1-x/x]+x√[1-x/x]+c
∫(x^2乘以sinx)/(1+x^2) dx 区间-π/2 到π/2
=∫ sinx dx - ∫ sinx/(1+x^2) dx
因为积分上限与下限是对称的,且被积分的函数是奇函数,所以积分后的函数值为0
=0