∫[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"(x)/f''^3(x)]dx如题∫[f(x)/f''(x)-f^2(x)f"(x)/f
∫[f(x)/f'(x)-f^2(x)f"(x)/f'^3(x)]dx 如题
∫[f(x)/f'(x)-f^2(x)f"(x)/f'^3(x)]dx 如题
∫[f(x)/f'(x)-f^2(x)f"(x)/f'^3(x)]dx 如题
[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”.
不定积分的主要方法就是凑微分法。
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