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人気のある微分積分 >

integral of (x^2)/(sqrt(x-x^2))

解

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

解

\frac{1}{16} (6 arcsin (2 x - 1) - sin (2 arcsin (2 x - 1)) - 16 x (- x + 1)) + C

解答ステップ

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

平方完成する

x - x^{2} : - (x - \frac{1}{2})^{2} + \frac{1}{4}

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

u置換積分法を適用する

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

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

拡張

\frac{( 2 u + 1 ) ^{2}}{- 4 u ^{2} + 1} : \frac{4 u ^{2}}{- 4 u ^{2} + 1} + \frac{4 u}{- 4 u ^{2} + 1} + \frac{1}{- 4 u ^{2} + 1}

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

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{4 u ^{2}}{- 4 u ^{2} + 1} d u = \frac{1}{4} (arcsin (2 u) - \frac{1}{2} sin (2 arcsin (2 u)))

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

\int \frac{1}{- 4 u ^{2} + 1} d u = \frac{1}{2} arcsin (2 u)

= \frac{1}{2} (\frac{1}{4} (arcsin (2 u) - \frac{1}{2} sin (2 arcsin (2 u))) - - 4 u^{2} + 1 + \frac{1}{2} arcsin (2 u))

代用を戻す

u = x - \frac{1}{2}

= \frac{1}{2} \frac{1}{4} (arcsin (2 (x - \frac{1}{2})) - \frac{1}{2} sin (2 arcsin (2 (x - \frac{1}{2})))) - - 4 (x - \frac{1}{2})^{2} + 1 + \frac{1}{2} arcsin (2 (x - \frac{1}{2}))

簡素化

\frac{1}{2} \frac{1}{4} (arcsin (2 (x - \frac{1}{2})) - \frac{1}{2} sin (2 arcsin (2 (x - \frac{1}{2})))) - - 4 (x - \frac{1}{2})^{2} + 1 + \frac{1}{2} arcsin (2 (x - \frac{1}{2})) : \frac{1}{16} (6 arcsin (2 x - 1) - sin (2 arcsin (2 x - 1)) - 16 x (- x + 1))

= \frac{1}{16} (6 arcsin (2 x - 1) - sin (2 arcsin (2 x - 1)) - 16 x (- x + 1))

定数を解答に追加する

= \frac{1}{16} (6 arcsin (2 x - 1) - sin (2 arcsin (2 x - 1)) - 16 x (- x + 1)) + C

グラフ

人気の例

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