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

解

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

解

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

+1

十進法表記

0.02634 \dots

解答ステップ

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

\frac{1}{2 x ^{2} - 2} = \frac{1}{2} \frac{1}{x ^{2} - 1}

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

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

因数

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

= \frac{1}{2} \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

= \frac{1}{2} (- \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}

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

簡素化

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

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

限度を計算する:

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

= - \frac{1}{2} \cdot \frac{ln ( 6 ) - 2 ln ( 2 ) - ln ( 5 ) + ln ( 3 )}{2}

簡素化

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

グラフ

人気の例

integral from 0 to infinity of e^{-sx}x

\int_{0}^{\infty} e^{- s x} x d x

integral from 1 to 4 of 3x^2+5

\int_{1}^{4} 3 x^{2} + 5 d x

integral from 1 to e^6 of x^3ln(x)

\int_{1}^{e^{6}} x^{3} ln (x) d x

integral from 0 to a of (x^4+5)^3x^3

\int_{0}^{a} (x^{4} + 5)^{3} x^{3} d x

integral from 0 to 4 of |3x-9|

\int_{0}^{4} ∣ 3 x - 9 ∣ d x