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人気のある微分積分 >

integral of 9/(5xsqrt(49-25x^2))

解

\int \frac{9}{5 x 49 - 25 x ^{2}} d x

解

- \frac{9}{70} (ln \frac{49 - 25 x ^{2}}{7} + 1 - ln \frac{49 - 25 x ^{2}}{7} - 1) + C

解答ステップ

\int \frac{9}{5 x 49 - 25 x ^{2}} d x

定数を除く:

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

= \frac{9}{5} \cdot \int \frac{1}{x 49 - 25 x ^{2}} d x

u置換積分法を適用する

= \frac{9}{5} \cdot \int \frac{1}{2 u 49 - 25 u} d u

定数を除く:

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

= \frac{9}{5} \cdot \frac{1}{2} \cdot \int \frac{1}{u 49 - 25 u} d u

u置換積分法を適用する

= \frac{9}{5} \cdot \frac{1}{2} \cdot \int - \frac{2}{- v ^{2} + 49} d v

定数を除く:

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

= \frac{9}{5} \cdot \frac{1}{2} (- 2 \cdot \int \frac{1}{- v ^{2} + 49} d v)

置換積分法を適用する

= \frac{9}{5} \cdot \frac{1}{2} (- 2 \cdot \int \frac{1}{7 ( - w ^{2} + 1 )} d w)

定数を除く:

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

= \frac{9}{5} \cdot \frac{1}{2} (- 2 \cdot \frac{1}{7} \cdot \int \frac{1}{- w ^{2} + 1} d w)

共通積分を使用する:

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

= \frac{9}{5} \cdot \frac{1}{2} (- 2 \cdot \frac{1}{7} (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

代用を戻す

= \frac{9}{5} \cdot \frac{1}{2} - 2 \cdot \frac{1}{7} \frac{ln \frac{49 - 25 x ^{2}}{7} + 1}{2} - \frac{ln \frac{49 - 25 x ^{2}}{7} - 1}{2}

簡素化

\frac{9}{5} \cdot \frac{1}{2} - 2 \cdot \frac{1}{7} \frac{ln \frac{49 - 25 x ^{2}}{7} + 1}{2} - \frac{ln \frac{49 - 25 x ^{2}}{7} - 1}{2} : - \frac{9}{70} (ln \frac{49 - 25 x ^{2}}{7} + 1 - ln \frac{49 - 25 x ^{2}}{7} - 1)

= - \frac{9}{70} (ln \frac{49 - 25 x ^{2}}{7} + 1 - ln \frac{49 - 25 x ^{2}}{7} - 1)

定数を解答に追加する

= - \frac{9}{70} (ln \frac{49 - 25 x ^{2}}{7} + 1 - ln \frac{49 - 25 x ^{2}}{7} - 1) + C

グラフ

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