integral from 1 to e of (ln(x))/(x^3)

解

\int_{1}^{e} \frac{ln ( x )}{x ^{3}} d x

- \frac{3}{4 e ^{2}} + \frac{1}{4}

十進法表記

0.14849 \dots

解答ステップ

\int_{1}^{e} \frac{ln ( x )}{x ^{3}} d x

部分積分を適用する

= [\frac{- ln ( x ) - 2 x ^{2} \cdot \int - \frac{1}{2 x ^{3}} d x}{2 x ^{2}}]_{1}^{e}

\int - \frac{1}{2 x ^{3}} d x = \frac{1}{4 x ^{2}}

= [\frac{- ln ( x ) - 2 x ^{2} \frac{1}{4 x ^{2}}}{2 x ^{2}}]_{1}^{e}

\frac{- ln ( x ) - 2 x ^{2} \frac{1}{4 x ^{2}}}{2 x ^{2}} = \frac{- 2 ln ( x ) - 1}{4 x ^{2}}

= [\frac{- 2 ln ( x ) - 1}{4 x ^{2}}]_{1}^{e}

限度を計算する:

- \frac{3}{4 e ^{2}} + \frac{1}{4}

= - \frac{3}{4 e ^{2}} + \frac{1}{4}