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人気のある微分積分 >

integral from 4 to 5 of 4/(x^2-1)

解

\int_{4}^{5} \frac{4}{x ^{2} - 1} d x

解

- 2 (ln (\frac{9}{2}) - ln (5))

+1

十進法表記

0.21072 \dots

解答ステップ

\int_{4}^{5} \frac{4}{x ^{2} - 1} d x

定数を除く:

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

= 4 \cdot \int_{4}^{5} \frac{1}{x ^{2} - 1} d x

因数

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

= 4 \cdot \int_{4}^{5} \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

= 4 (- \int_{4}^{5} \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}

= 4 (- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{4}^{5})

簡素化

4 (- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{4}^{5}) : - 4 [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{4}^{5}

= - 4 [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{4}^{5}

限度を計算する:

\frac{ln ( 6 ) - 2 ln ( 2 ) - ln ( 5 ) + ln ( 3 )}{2}

= - 4 \cdot \frac{ln ( 6 ) - 2 ln ( 2 ) - ln ( 5 ) + ln ( 3 )}{2}

簡素化

= - 2 (ln (\frac{9}{2}) - ln (5))

グラフ

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