请解微分 x^2/(1-cosx)

x^2/(1-cosx)=x^2/(2(sinx/2)^2)用等价无穷小代换 原式=x^2/(x^2/2)=2

d[x²/(1-cosx)]
=[(1-cosx)d(x²)-x²d(1-cosx)]/(1-cosx)²
=[2(1-cosx)-x²sinx]dx/(1-cosx)²
=[(2-2cosx-x²sinx)/(1-cosx)²]dx

[(x^2)/(1+cosx)]dx={(x^2)'/(1+cosx)+(x^2)*[1/(1+cosx)]'}dx={2x/(1+cosx)+(x^2*sinx)/[(1+cosx)^2]}dx=差不多了

(2x*(1-cosx)-x^2*sinx)/(1-cosx)^2