∫[f(x)/f’(x)-f^2(x)f’’(x)/f’^3(x)]dx
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∫[f(x)/f’(x)-f^2(x)f’’(x)/f’^3(x)]dx
∫[f(x)/f’(x)-f^2(x)f’’(x)/f’^3(x)]dx
∫[f(x)/f’(x)-f^2(x)f’’(x)/f’^3(x)]dx
第一步通分.之后[f(x)/f'(x)]df(x)/f'(x)=1/2[fx/f'x]^2 我写的可能简单勒点.不过应该能看懂吧.嘿嘿.加油
∫[f(x)/f’(x)-f^2(x)f’’(x)/f’^3(x)]dx=∫{[f(x)f’^2(x)-f^2(x)f’’(x)]/f’^3(x)}dx=∫1/2*{[2f(x)f’^3(x)-2f^2(x)f’(x)f’’(x)]/f’^4(x)}dx=1/2*(f^2(x)/f’(x))。
∫[f(x)/f’(x)-f^2(x)f’’(x)/f’^3(x)]dx
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