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人気のある微分積分 >

integral of 1/(x^2-8)

解

\int \frac{1}{x ^{2} - 8} d x

解

- \frac{1}{4 2} (ln \frac{x}{2 2} + 1 - ln \frac{x}{2 2} - 1) + C

解答ステップ

\int \frac{1}{x ^{2} - 8} d x

置換積分法を適用する

= \int \frac{1}{2 2 ( u ^{2} - 1 )} d u

定数を除く:

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

= \frac{1}{2 2} \cdot \int \frac{1}{u ^{2} - 1} d u

因数

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

= \frac{1}{2 2} \cdot \int \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

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

= \frac{1}{2 2} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

代用を戻す

u = \frac{x}{2 2}

= \frac{1}{2 2} - \frac{ln \frac{x}{2 2} + 1}{2} - \frac{ln \frac{x}{2 2} - 1}{2}

簡素化

\frac{1}{2 2} - \frac{ln \frac{x}{2 2} + 1}{2} - \frac{ln \frac{x}{2 2} - 1}{2} : - \frac{1}{4 2} (ln \frac{x}{2 2} + 1 - ln \frac{x}{2 2} - 1)

= - \frac{1}{4 2} (ln \frac{x}{2 2} + 1 - ln \frac{x}{2 2} - 1)

定数を解答に追加する

= - \frac{1}{4 2} (ln \frac{x}{2 2} + 1 - ln \frac{x}{2 2} - 1) + C

グラフ

人気の例

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\int \frac{5 x + 1}{25 x ^{2} + 60 x - 13} d x

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