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人気のある微分積分 >

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

解

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

解

- ln (2) + ln (3)

+1

十進法表記

0.40546 \dots

解答ステップ

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

定数を除く:

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

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

因数

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

= 2 \cdot \int_{2}^{3} \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

= 2 (- \int_{2}^{3} \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}

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

簡素化

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

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

限度を計算する:

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

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

簡素化

= - ln (2) + ln (3)

グラフ

人気の例

integral of x^2-2x

\int x^{2} - 2 x d x

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

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

f (x) = x^{3} cos (x)

(\partial)/(\partial x)((4x+3)/(4y+1))

\frac{\partial}{\partial x} (\frac{4 x + 3}{4 y + 1})

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\frac{d}{d x} (\frac{ln ( x )}{x ^{\frac{5}{4}}})