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人気のある微分積分 >

integral of (x^2)/(sqrt(1+x-x^2))

解

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

解

\frac{1}{16} (14 arcsin (\frac{2 x - 1}{5}) - 5 sin (2 arcsin (\frac{2 x - 1}{5})) - 16 - x^{2} + x + 1) + C

解答ステップ

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

平方完成する

1 + x - x^{2} : - (x - \frac{1}{2})^{2} + \frac{5}{4}

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

u置換積分法を適用する

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

拡張

\frac{( 2 u + 1 ) ^{2}}{- 4 u ^{2} + 5} : \frac{4 u ^{2}}{- 4 u ^{2} + 5} + \frac{4 u}{- 4 u ^{2} + 5} + \frac{1}{- 4 u ^{2} + 5}

= \frac{1}{2} \cdot \int \frac{4 u ^{2}}{- 4 u ^{2} + 5} + \frac{4 u}{- 4 u ^{2} + 5} + \frac{1}{- 4 u ^{2} + 5} 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} + 5} d u + \int \frac{4 u}{- 4 u ^{2} + 5} d u + \int \frac{1}{- 4 u ^{2} + 5} d u)

\int \frac{4 u ^{2}}{- 4 u ^{2} + 5} d u = \frac{5}{4} (arcsin (\frac{2}{5} u) - \frac{1}{2} sin (2 arcsin (\frac{2}{5} u)))

\int \frac{4 u}{- 4 u ^{2} + 5} d u = - - 4 u^{2} + 5

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

= \frac{1}{2} (\frac{5}{4} (arcsin (\frac{2}{5} u) - \frac{1}{2} sin (2 arcsin (\frac{2}{5} u))) - - 4 u^{2} + 5 + \frac{1}{2} arcsin (\frac{2}{5} u))

代用を戻す

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

= \frac{1}{2} \frac{5}{4} (arcsin (\frac{2}{5} (x - \frac{1}{2})) - \frac{1}{2} sin (2 arcsin (\frac{2}{5} (x - \frac{1}{2})))) - - 4 (x - \frac{1}{2})^{2} + 5 + \frac{1}{2} arcsin (\frac{2}{5} (x - \frac{1}{2}))

簡素化

\frac{1}{2} \frac{5}{4} (arcsin (\frac{2}{5} (x - \frac{1}{2})) - \frac{1}{2} sin (2 arcsin (\frac{2}{5} (x - \frac{1}{2})))) - - 4 (x - \frac{1}{2})^{2} + 5 + \frac{1}{2} arcsin (\frac{2}{5} (x - \frac{1}{2})) : \frac{1}{16} (14 arcsin (\frac{2 x - 1}{5}) - 5 sin (2 arcsin (\frac{2 x - 1}{5})) - 16 - x^{2} + x + 1)

= \frac{1}{16} (14 arcsin (\frac{2 x - 1}{5}) - 5 sin (2 arcsin (\frac{2 x - 1}{5})) - 16 - x^{2} + x + 1)

定数を解答に追加する

= \frac{1}{16} (14 arcsin (\frac{2 x - 1}{5}) - 5 sin (2 arcsin (\frac{2 x - 1}{5})) - 16 - x^{2} + x + 1) + C

グラフ

人気の例

integral of (x^3+1)/(sqrt(x^4+4x))

\int \frac{x ^{3} + 1}{x ^{4} + 4 x} d x

integral of (4x^2)/(1+x^2)

\int \frac{4 x ^{2}}{1 + x ^{2}} d x

integral of (7-5x)sin(3-2x)

\int (7 - 5 x) sin (3 - 2 x) d x

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

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

integral of 1/(sqrt(x+1)-\sqrt{x-1)}

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