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