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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

拡張

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

= \int \frac{2 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

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

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of 3x^2*ln(x^2+3)

\int 3 x^{2} \cdot ln (x^{2} + 3) d x

integral of 1/(cos^2(x)sqrt(tan(x)))

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

integral of-2x^4e^{x^5}

\int - 2 x^{4} e^{x^{5}} d x

integral of (x^2)/(sqrt(x-3))

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

integral of (3x^2-7x-2)/(x^3-x)

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