求∫(0~e^(－1)）ln(x+1)dx的解題過程及結(jié)果
求∫(0~e^(－1)）ln(x+1)dx的解題過程及結(jié)果
數(shù)學(xué)人氣：944 ℃時(shí)間：2020-01-31 18:40:22
優(yōu)質(zhì)解答
∫(0-->e^(-1)) ln(x + 1) dx= xln(x + 1) - ∫(0-->e^(-1)) x dln(x + 1) e^(-1)) x/(x + 1) dx= (1/e)ln(1 + 1/e) - ∫(0-->e^(-1)) [1 - 1/(x + 1)] dx= (1/e)[ln(e + 1) - lne] - [x - ln(x + 1)] |(0-->e^(-1))...
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