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人気のある微分積分 >

integral of (-4x-1)/(7(x^2-x+1))

解

\int \frac{- 4 x - 1}{7 ( x ^{2} - x + 1 )} d x

解

\frac{1}{7 3} (- 23 ln 4 x^{2} - 4 x + 4 - 6 arctan (\frac{2 x - 1}{3})) + C

解答ステップ

\int \frac{- 4 x - 1}{7 ( x ^{2} - x + 1 )} d x

定数を除く:

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

= \frac{1}{7} \cdot \int \frac{- 4 x - 1}{x ^{2} - x + 1} d x

拡張

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

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

\int \frac{4 x}{x ^{2} - x + 1} d x = 8 (\frac{1}{4} ln 4 x^{2} - 4 x + 4 + \frac{1}{2 3} arctan (\frac{2 x - 1}{3}))

\int \frac{1}{x ^{2} - x + 1} d x = \frac{2}{3} arctan (\frac{2 x - 1}{3})

= \frac{1}{7} (- 8 (\frac{1}{4} ln 4 x^{2} - 4 x + 4 + \frac{1}{2 3} arctan (\frac{2 x - 1}{3})) - \frac{2}{3} arctan (\frac{2 x - 1}{3}))

簡素化

\frac{1}{7} (- 8 (\frac{1}{4} ln 4 x^{2} - 4 x + 4 + \frac{1}{2 3} arctan (\frac{2 x - 1}{3})) - \frac{2}{3} arctan (\frac{2 x - 1}{3})) : \frac{1}{7 3} (- 23 ln 4 x^{2} - 4 x + 4 - 6 arctan (\frac{2 x - 1}{3}))

= \frac{1}{7 3} (- 23 ln 4 x^{2} - 4 x + 4 - 6 arctan (\frac{2 x - 1}{3}))

定数を解答に追加する

= \frac{1}{7 3} (- 23 ln 4 x^{2} - 4 x + 4 - 6 arctan (\frac{2 x - 1}{3})) + C

グラフ

人気の例

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