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人気のある微分積分 >

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

解

\int_{- 1}^{- 2} \frac{e ^{x}}{1 - e ^{x}} d x

解

- ln (\frac{e ^{2} - 1}{e ^{2} - e})

+1

十進法表記

- 0.31326 \dots

解答ステップ

\int_{- 1}^{- 2} \frac{e ^{x}}{1 - e ^{x}} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 2}^{- 1} \frac{e ^{x}}{1 - e ^{x}} d x

u置換積分法を適用する

= - \int_{1 - \frac{1}{e ^{2}}}^{1 - \frac{1}{e}} - \frac{1}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - (- \int_{1 - \frac{1}{e}}^{1 - \frac{1}{e ^{2}}} - \frac{1}{u} d u)

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- (- \int_{1 - \frac{1}{e}}^{1 - \frac{1}{e ^{2}}} \frac{1}{u} d u))

共通積分を使用する:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

簡素化

= - [ln ∣ u ∣]_{1 - \frac{1}{e}}^{1 - \frac{1}{e ^{2}}}

限度を計算する:

ln (1 - \frac{1}{e ^{2}}) - ln (1 - \frac{1}{e})

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

簡素化

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

グラフ

人気の例

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