xf(x)-f'(x)x^2=∫f(t)dt 求导
f(x)+xf'(x)-f''(x)x^2-2xf'(x)=f(x)
f''(x)x+f'(x)=0
令f'(x)=p
xdp/dx=-p
dp/p=-dx/x
lnp=-lnx+C
p=e^(-lnx+C1)=C2e^(-lnx)=C2/x
df(x)/dx=C2/x
df(x)=C2/x dx
f(x)=C2lnx+C3