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人気のある微分積分 >

integral from 0 to 1 of (16)/(x^2-16)

解

\int_{0}^{1} \frac{16}{x ^{2} - 16} d x

解

- 2 (ln (\frac{5}{4}) - ln (\frac{3}{4}))

+1

十進法表記

- 1.02165 \dots

解答ステップ

\int_{0}^{1} \frac{16}{x ^{2} - 16} d x

定数を除く:

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

= 16 \cdot \int_{0}^{1} \frac{1}{x ^{2} - 16} d x

置換積分法を適用する

= 16 \cdot \int_{0}^{\frac{1}{4}} \frac{1}{4 ( u ^{2} - 1 )} d u

定数を除く:

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

= 16 \cdot \frac{1}{4} \cdot \int_{0}^{\frac{1}{4}} \frac{1}{u ^{2} - 1} d u

因数

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

= 16 \cdot \frac{1}{4} \cdot \int_{0}^{\frac{1}{4}} \frac{1}{- ( - u ^{2} + 1 )} d u

定数を除く:

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

= 16 \cdot \frac{1}{4} (- \int_{0}^{\frac{1}{4}} \frac{1}{- u ^{2} + 1} d u)

共通積分を使用する:

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

= 16 \cdot \frac{1}{4} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{4}})

簡素化

16 \cdot \frac{1}{4} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{4}}) : - 4 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{\frac{1}{4}}

= - 4 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{\frac{1}{4}}

限度を計算する:

\frac{ln ( \frac{5}{4} ) - ln ( \frac{3}{4} )}{2}

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

簡素化

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

グラフ

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