问题描述: ∫[f(x)/f'(x)-f^2(x)f"(x)/f'^3(x)]dx 如题 1个回答 分类:综合 2014-11-10 问题解答: 我来补答 [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”. 展开全文阅读