∵∫[f(u)-f(1/x)]du =∫f(u)du-∫f(1/x)du
=∫f(u)du-(1-1/x)f(1/x)
∴[∫[f(u)-f(1/x)]du]'=[∫f(u)du-(1-1/x)f(1/x)]'
=[∫f(u)du]'-[(1-1/x)f(1/x)]'
=-f(1/x)(1/x)'-[f(1/x)(1-1/x)'+(1-1/x)f'(1/x)(1/x)']
=f(1/x)/x²-f(1/x)/x²+(1-1/x)f'(1/x)/x²
=f'(1/x)(x-1)/x³.
∫1/x到1 [f(u)-f(1/x)]du 求導(dǎo)
∫1/x到1 [f(u)-f(1/x)]du 求導(dǎo)
其他人氣：764 ℃時(shí)間：2020-05-02 18:59:49
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