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人気のある微分積分 >

integral of 9/(x^2-16)

解

\int \frac{9}{x ^{2} - 16} d x

解

- \frac{9}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1) + C

解答ステップ

\int \frac{9}{x ^{2} - 16} d x

定数を除く:

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

= 9 \cdot \int \frac{1}{x ^{2} - 16} d x

置換積分法を適用する

= 9 \cdot \int \frac{1}{4 ( u ^{2} - 1 )} d u

定数を除く:

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

= 9 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{2} - 1} d u

因数

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

= 9 \cdot \frac{1}{4} \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

= 9 \cdot \frac{1}{4} (- \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}

= 9 \cdot \frac{1}{4} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

代用を戻す

u = \frac{x}{4}

= 9 \cdot \frac{1}{4} (- (\frac{ln \frac{x}{4} + 1}{2} - \frac{ln \frac{x}{4} - 1}{2}))

簡素化

9 \cdot \frac{1}{4} (- (\frac{ln \frac{x}{4} + 1}{2} - \frac{ln \frac{x}{4} - 1}{2})) : - \frac{9}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1)

= - \frac{9}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1)

定数を解答に追加する

= - \frac{9}{8} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1) + C

グラフ

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