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人気のある微分積分 >

integral of 1/(xsqrt(2-x^2))

解

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

解

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

解答ステップ

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

u置換積分法を適用する

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

定数を除く:

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

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

置換積分法を適用する

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

定数を除く:

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

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

共通積分を使用する:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

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

代用を戻す

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

簡素化

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

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

定数を解答に追加する

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

グラフ

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