integral from e^{-t} to e^{1-t of}ln(x)

解答

\int_{e^{- t}}^{e^{1 - t}} ln (x) d x

- t e^{- t + 1} + t e^{- t} + e^{- t}

求解步骤

\int_{e^{- t}}^{e^{1 - t}} ln (x) d x

使用分布积分法

= [x ln (x) - \int 1 d x]_{e^{- t}}^{e^{1 - t}}

\int 1 d x = x

= [x ln (x) - x]_{e^{- t}}^{e^{1 - t}}

带入上下限计算后得:

- t e^{- t + 1} + t e^{- t} + e^{- t}

= - t e^{- t + 1} + t e^{- t} + e^{- t}

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