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人気のある微分積分 >

integral of 2/(c^2-x^2)

解

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

解

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

解答ステップ

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

定数を除く:

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

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

因数

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

= 2 \cdot \int \frac{1}{- ( - c ^{2} + x ^{2} )} d x

定数を除く:

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

= 2 (- \int \frac{1}{- c ^{2} + x ^{2}} d x)

置換積分法を適用する

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

定数を除く:

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

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

因数

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

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

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

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

代用を戻す

u = \frac{x}{c}

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

簡素化

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

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

定数を解答に追加する

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

グラフ

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