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人気のある微分積分 >

integral from-2 to-3 of 1/(x^2-1)

解

\int_{- 2}^{- 3} \frac{1}{x ^{2} - 1} d x

解

\frac{- ln ( 3 ) + ln ( 2 )}{2}

+1

十進法表記

- 0.20273 \dots

解答ステップ

\int_{- 2}^{- 3} \frac{1}{x ^{2} - 1} d x

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

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

因数

x^{2} - 1 : - (- x^{2} + 1)

= - \int_{- 3}^{- 2} \frac{1}{- ( - x ^{2} + 1 )} d x

定数を除く:

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

= - (- \int_{- 3}^{- 2} \frac{1}{- x ^{2} + 1} d x)

共通積分を使用する:

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

= - (- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{- 3}^{- 2})

簡素化

- (- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{- 3}^{- 2}) : [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{- 3}^{- 2}

= [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{- 3}^{- 2}

限度を計算する:

\frac{- ln ( 3 ) + ln ( 2 )}{2}

= \frac{- ln ( 3 ) + ln ( 2 )}{2}

グラフ

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