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人気のある微分積分 >

integral from 1 to e of ((ln(x))/x)^2

解

\int_{1}^{e} (\frac{ln ( x )}{x})^{2} d x

解

- \frac{5}{e} + 2

+1

十進法表記

0.16060 \dots

解答ステップ

\int_{1}^{e} (\frac{ln ( x )}{x})^{2} d x

指数の規則を適用する:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \int_{1}^{e} \frac{ln ^{2} ( x )}{x ^{2}} d x

部分積分を適用する

= [\frac{- ln ^{2} ( x ) - x \cdot \int - \frac{2 l n ( x )}{x ^{2}} d x}{x}]_{1}^{e}

\int - \frac{2 ln ( x )}{x ^{2}} d x = - 2 (- \frac{ln ( x )}{x} - \frac{1}{x})

= \frac{- ln ^{2} ( x ) - x ( - 2 ( - \frac{l n ( x )}{x} - \frac{1}{x} ) )}{x}_{1}^{e}

簡素化

\frac{- ln ^{2} ( x ) - x ( - 2 ( - \frac{l n ( x )}{x} - \frac{1}{x} ) )}{x}_{1}^{e} : [\frac{- ln ^{2} ( x ) - 2 ln ( x ) - 2}{x}]_{1}^{e}

= [\frac{- ln ^{2} ( x ) - 2 ln ( x ) - 2}{x}]_{1}^{e}

限度を計算する:

- \frac{5}{e} + 2

= - \frac{5}{e} + 2

グラフ

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