∫ x^2. cos(1/x) dx
=(1/3)∫ cos(1/x) dx^3
=(1/3)x^3.cos(1/x)-(1/3)∫ xsin(1/x) dx
=(1/3)x^3.cos(1/x)-(1/6)∫ sin(1/x) dx^2
=(1/3)x^3.cos(1/x)-(1/6)x^2.sin(1/x)-(1/6)∫ cos(1/x) dx
=(1/3)x^3.cos(1/x)-(1/6)x^2.sin(1/x)-(1/12)[xcos(1/x) +(1/x^2)sin(1/x) ] +C
consider
∫ cos(1/x) dx
=xcos(1/x) -∫ (1/x)sin(1/x) dx
=xcos(1/x) +∫ sin(1/x) d(1/x^2)
=xcos(1/x) +(1/x^2)sin(1/x) -∫ cos(1/x) dx
2∫ cos(1/x) dx =xcos(1/x) +(1/x^2)sin(1/x)
∫ cos(1/x) dx
=(1/2)[xcos(1/x) +(1/x^2)sin(1/x) ] +C
追问∫ (1/x)sin(1/x) dx是怎么转化为∫ sin(1/x) d(1/x^2)的呀?不应该是∫ sin(1/x) dlnx吗