∫[f(x)/f'(x)-f^2(x)f"(x)/f'^3(x)]dx 如题
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∫[f(x)/f'(x)-f^2(x)f"(x)/f'^3(x)]dx 如题
![∫[f(x)/f'(x)-f^2(x)f](/uploads/image/z/4193737-25-7.jpg?t=%E2%88%AB%5Bf%28x%29%2Ff%27%28x%29-f%5E2%28x%29f%22%28x%29%2Ff%27%5E3%28x%29%5Ddx+%E5%A6%82%E9%A2%98)
[f(x)/f '(x)]'=[f '²(x)-f(x)f ''(x)]/f '²(x)
=1-f(x)f ''(x)/f '²(x)
因此题目中的被积函数为:
[f(x)/f'(x)-f^2(x)f"(x)/f'^3(x)]
=[f(x)/f '(x)][1-f(x)f ''(x)/f '²(x)]
=[f(x)/f '(x)][f(x)/f '(x)]'
因此:原式=∫ [f(x)/f '(x)][f(x)/f '(x)]' dx
=∫ [f(x)/f '(x)][f(x)/f '(x)]' dx
=∫ [f(x)/f '(x)] d[f(x)/f '(x)]
=(1/2)[f(x)/f '(x)]² + C
若有不懂请追问,如果解决问题请点下面的“选为满意答案”.
=1-f(x)f ''(x)/f '²(x)
因此题目中的被积函数为:
[f(x)/f'(x)-f^2(x)f"(x)/f'^3(x)]
=[f(x)/f '(x)][1-f(x)f ''(x)/f '²(x)]
=[f(x)/f '(x)][f(x)/f '(x)]'
因此:原式=∫ [f(x)/f '(x)][f(x)/f '(x)]' dx
=∫ [f(x)/f '(x)][f(x)/f '(x)]' dx
=∫ [f(x)/f '(x)] d[f(x)/f '(x)]
=(1/2)[f(x)/f '(x)]² + C
若有不懂请追问,如果解决问题请点下面的“选为满意答案”.
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