ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral from 0 to 1 of (36)/(x^2-81)

解

\int_{0}^{1} \frac{36}{x ^{2} - 81} d x

解

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

+1

十進法表記

- 0.44628 \dots

解答ステップ

\int_{0}^{1} \frac{36}{x ^{2} - 81} d x

定数を除く:

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

= 36 \cdot \int_{0}^{1} \frac{1}{x ^{2} - 81} d x

置換積分法を適用する

= 36 \cdot \int_{0}^{\frac{1}{9}} \frac{1}{9 ( u ^{2} - 1 )} d u

定数を除く:

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

= 36 \cdot \frac{1}{9} \cdot \int_{0}^{\frac{1}{9}} \frac{1}{u ^{2} - 1} d u

因数

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

= 36 \cdot \frac{1}{9} \cdot \int_{0}^{\frac{1}{9}} \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

= 36 \cdot \frac{1}{9} (- \int_{0}^{\frac{1}{9}} \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}

= 36 \cdot \frac{1}{9} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{9}})

簡素化

36 \cdot \frac{1}{9} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{9}}) : - 4 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{\frac{1}{9}}

= - 4 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{\frac{1}{9}}

限度を計算する:

\frac{ln ( \frac{10}{9} ) - ln ( \frac{8}{9} )}{2}

= - 4 \cdot \frac{ln ( \frac{10}{9} ) - ln ( \frac{8}{9} )}{2}

簡素化

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

グラフ

人気の例

integral from 0 to pi of 6xsin(x)

\int_{0}^{π} 6 x sin (x) d x

integral from 0 to 2 of 1/((x-1)^4)

\int_{0}^{2} \frac{1}{( x - 1 ) ^{4}} d x

integral from 0 to 1 of (-t+1)e^{-st}

\int_{0}^{1} (- t + 1) e^{- s t} d t

integral from 0 to 1/2 of sqrt(9-36x^2)

\int_{0}^{\frac{1}{2}} 9 - 36 x^{2} d x

integral from 1 to x of xy

\int_{1}^{x} x y d y