∫ 1/√[（4x^2+9)^3] dx
解設x=3/2*tant,則原式化為
∫ 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,則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 區(qū)間-π/2 到π/2
=∫ sinx dx - ∫ sinx/(1+x^2) dx
因為積分上限與下限是對稱的,且被積分的函數(shù)是奇函數(shù),所以積分后的函數(shù)值為0
=0