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人気のある微分積分 >

integral of 1/(25-4x^2)

解

\int \frac{1}{25 - 4 x ^{2}} d x

解

\frac{1}{20} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1) + C

解答ステップ

\int \frac{1}{25 - 4 x ^{2}} d x

置換積分法を適用する

= \int \frac{1}{10 ( - 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}{10} \cdot \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}{10} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

代用を戻す

u = \frac{2}{5} x

= \frac{1}{10} (\frac{ln \frac{2}{5} x + 1}{2} - \frac{ln \frac{2}{5} x - 1}{2})

簡素化

\frac{1}{10} (\frac{ln \frac{2}{5} x + 1}{2} - \frac{ln \frac{2}{5} x - 1}{2}) : \frac{1}{20} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1)

= \frac{1}{20} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1)

定数を解答に追加する

= \frac{1}{20} (ln \frac{2}{5} x + 1 - ln \frac{2}{5} x - 1) + C

グラフ

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