已知f'(sin^2x)=cos2x+tan^2x,当0

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已知f'(sin^2x)=cos2x+tan^2x,当0已知f'(sin^2x)=cos2x+tan^2x,当0已知f'(sin^2x)=cos2x+tan^2x,当0f

已知f'(sin^2x)=cos2x+tan^2x,当0
已知f'(sin^2x)=cos2x+tan^2x,当0

已知f'(sin^2x)=cos2x+tan^2x,当0
f'(sin^2x)=1-2sin^2x+sin^2x/(1-sin^2x) f'(x)=1-2x+x/(1-x) f(x)=∫[1-2x+x/(1-x)]dx =x-x^2+∫x/(1-x)dx =x-x^2-∫-x/(1-x)dx =x-x^2-∫[1+1/(x-1)]dx =x-x^2-x-ln(1-x)+C =-x^2-ln(1-x)+C 追问:=x-x^2-∫[1+1/(x-1)]dx =x-x^2-x-ln(1-x)+C ln(1-x)的ln前是不是应该是+,为什么是-?