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人気のある微分積分 >

integral of (2x-8)/(sqrt(1-x-x^2))

解

\int \frac{2 x - 8}{1 - x - x ^{2}} d x

解

- 2 - x^{2} - x + 1 - 9 arcsin (\frac{1}{5} (2 x + 1)) + C

解答ステップ

\int \frac{2 x - 8}{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{2 x - 8}{- ( x + \frac{1}{2} ) ^{2} + \frac{5}{4}} d x

u置換積分法を適用する

= \int \frac{2 ( 2 u - 9 )}{- 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

= 2 \cdot \int \frac{2 u - 9}{- 4 u ^{2} + 5} d u

拡張

\frac{2 u - 9}{- 4 u ^{2} + 5} : \frac{2 u}{- 4 u ^{2} + 5} - \frac{9}{- 4 u ^{2} + 5}

= 2 \cdot \int \frac{2 u}{- 4 u ^{2} + 5} - \frac{9}{- 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

= 2 (\int \frac{2 u}{- 4 u ^{2} + 5} d u - \int \frac{9}{- 4 u ^{2} + 5} d u)

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

\int \frac{9}{- 4 u ^{2} + 5} d u = \frac{9}{2} arcsin (\frac{2}{5} u)

= 2 (- \frac{1}{2} - 4 u^{2} + 5 - \frac{9}{2} arcsin (\frac{2}{5} u))

代用を戻す

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

= 2 - \frac{1}{2} - 4 (x + \frac{1}{2})^{2} + 5 - \frac{9}{2} arcsin (\frac{2}{5} (x + \frac{1}{2}))

簡素化

2 - \frac{1}{2} - 4 (x + \frac{1}{2})^{2} + 5 - \frac{9}{2} arcsin (\frac{2}{5} (x + \frac{1}{2})) : - 2 - x^{2} - x + 1 - 9 arcsin (\frac{1}{5} (2 x + 1))

= - 2 - x^{2} - x + 1 - 9 arcsin (\frac{1}{5} (2 x + 1))

定数を解答に追加する

= - 2 - x^{2} - x + 1 - 9 arcsin (\frac{1}{5} (2 x + 1)) + C

グラフ

人気の例

(\partial)/(\partial y)(xln(xy-5))

\frac{\partial}{\partial y} (x ln (x y - 5))

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\int_{0}^{9} (\frac{1}{3} x - 2) d x

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\int u - 1 d u

integral of (tan(ln(4x+3)))/(4x+3)

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