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人気のある微分積分 >

integral of 1/(5x-x^2)

解

\int \frac{1}{5 x - x ^{2}} d x

解

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

解答ステップ

\int \frac{1}{5 x - x ^{2}} d x

平方完成する

5 x - x^{2} : - (x - \frac{5}{2})^{2} + \frac{25}{4}

= \int \frac{1}{- ( x - \frac{5}{2} ) ^{2} + \frac{25}{4}} d x

u置換積分法を適用する

= \int \frac{4}{- 4 u ^{2} + 25} d u

定数を除く:

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

= 4 \cdot \int \frac{1}{- 4 u ^{2} + 25} d u

置換積分法を適用する

= 4 \cdot \int \frac{1}{10 ( - v ^{2} + 1 )} d v

定数を除く:

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

= 4 \cdot \frac{1}{10} \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}

= 4 \cdot \frac{1}{10} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

代用を戻す

= 4 \cdot \frac{1}{10} (\frac{ln \frac{2}{5} ( x - \frac{5}{2} ) + 1}{2} - \frac{ln \frac{2}{5} ( x - \frac{5}{2} ) - 1}{2})

簡素化

4 \cdot \frac{1}{10} (\frac{ln \frac{2}{5} ( x - \frac{5}{2} ) + 1}{2} - \frac{ln \frac{2}{5} ( x - \frac{5}{2} ) - 1}{2}) : \frac{1}{5} (ln (\frac{2}{5} ∣ x ∣) - ln \frac{2}{5} x - 2)

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

定数を解答に追加する

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

グラフ

人気の例

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\int \frac{x ^{3} + x ^{2}}{x ^{2} + x - 2} d x

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\int (2 x^{5} - 20 x)^{3} (x^{4} - 2) d x

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\int \frac{cosh ( x )}{cosh ^{2} ( x ) - 1} d x

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integral of (12)/(sqrt(x+1))

\int \frac{12}{x + 1} d x