integral from 1 to e^4 of (ln(x))/(x^2)

解

\int_{1}^{e^{4}} \frac{ln ( x )}{x ^{2}} d x

- \frac{5}{e ^{4}} + 1

十進法表記

0.90842 \dots

解答ステップ

\int_{1}^{e^{4}} \frac{ln ( x )}{x ^{2}} d x

部分積分を適用する

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

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

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

x \frac{1}{x} = 1

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

限度を計算する:

- \frac{5}{e ^{4}} + 1

= - \frac{5}{e ^{4}} + 1