大学高数简单不定积分问题，来看看啦，这题好简单可我就是理解不了，谁来解释一下啊
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f(x)/f'(x)-f^2(x)f''(x)/f'^3(x)
=[f(x)f'^3(x)-f^2(x)f'(x)f''(x)]/f'^4(x)
令g(x)，使得[g(x)/f'^2(x)]'=[f(x)f'^3(x)-f^2(x)f'(x)f''(x)]/f'^4(x)
g'(x)f'^2(x)-g(x)2f'(x)f''(x)
g'(x)=f(x)f'(x)，g(x)=(1/2)*f^2(x)
所以f(x)/f'(x)-f^2(x)f''(x)/f'^3(x)=[(1/2)*f^2(x)/f'^2(x)]'
∫[f(x)/f'(x)-f^2(x)f''(x)/f'^3(x)]dx
=∫d[(1/2)*f^2(x)/f'^2(x)]
=(1/2)*[f(x)/f'(x)]^2+C，其中C是任意常数
=[f(x)f'^3(x)-f^2(x)f'(x)f''(x)]/f'^4(x)
令g(x)，使得[g(x)/f'^2(x)]'=[f(x)f'^3(x)-f^2(x)f'(x)f''(x)]/f'^4(x)
g'(x)f'^2(x)-g(x)2f'(x)f''(x)
g'(x)=f(x)f'(x)，g(x)=(1/2)*f^2(x)
所以f(x)/f'(x)-f^2(x)f''(x)/f'^3(x)=[(1/2)*f^2(x)/f'^2(x)]'
∫[f(x)/f'(x)-f^2(x)f''(x)/f'^3(x)]dx
=∫d[(1/2)*f^2(x)/f'^2(x)]
=(1/2)*[f(x)/f'(x)]^2+C，其中C是任意常数
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万分感谢😘😚😗😙
g(x)=1/2*f^2(x),这步怎么来的？它导数不是前面那个啊
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