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人気のある微分積分 >

integral from 0 to 1 of 3/(x^2-x-2)

解

\int_{0}^{1} \frac{3}{x ^{2} - x - 2} d x

解

- 2 ln (2)

+1

十進法表記

- 1.38629 \dots

解答ステップ

\int_{0}^{1} \frac{3}{x ^{2} - x - 2} d x

定数を除く:

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

= 3 \cdot \int_{0}^{1} \frac{1}{x ^{2} - x - 2} d x

平方完成する

x^{2} - x - 2 : (x - \frac{1}{2})^{2} - \frac{9}{4}

= 3 \cdot \int_{0}^{1} \frac{1}{( x - \frac{1}{2} ) ^{2} - \frac{9}{4}} d x

u置換積分法を適用する

= 3 \cdot \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{4}{4 u ^{2} - 9} d u

定数を除く:

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

= 3 \cdot 4 \cdot \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{1}{4 u ^{2} - 9} d u

置換積分法を適用する

= 3 \cdot 4 \cdot \int_{- \frac{1}{3}}^{\frac{1}{3}} \frac{1}{6 ( v ^{2} - 1 )} d v

定数を除く:

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

= 3 \cdot 4 \cdot \frac{1}{6} \cdot \int_{- \frac{1}{3}}^{\frac{1}{3}} \frac{1}{v ^{2} - 1} d v

因数

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

= 3 \cdot 4 \cdot \frac{1}{6} \cdot \int_{- \frac{1}{3}}^{\frac{1}{3}} \frac{1}{- ( - v ^{2} + 1 )} d v

定数を除く:

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

= 3 \cdot 4 \cdot \frac{1}{6} (- \int_{- \frac{1}{3}}^{\frac{1}{3}} \frac{1}{- v ^{2} + 1} d v)

共通積分を使用する:

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

= 3 \cdot 4 \cdot \frac{1}{6} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{- \frac{1}{3}}^{\frac{1}{3}})

簡素化

3 \cdot 4 \cdot \frac{1}{6} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{- \frac{1}{3}}^{\frac{1}{3}}) : - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{- \frac{1}{3}}^{\frac{1}{3}}

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

限度を計算する:

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

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

簡素化

= - 2 ln (2)

グラフ

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