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

解

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

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

解答ステップ

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

拡張

\frac{- 7 x + 4}{x ^{2} - x + 1} : - \frac{7 x}{x ^{2} - x + 1} + \frac{4}{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

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

\int \frac{7 x}{x ^{2} - x + 1} d x = 14 (\frac{1}{4} ln 4 x^{2} - 4 x + 4 + \frac{1}{2 3} arctan (\frac{2 x - 1}{3}))

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

= - 14 (\frac{1}{4} ln 4 x^{2} - 4 x + 4 + \frac{1}{2 3} arctan (\frac{2 x - 1}{3})) + \frac{8}{3} arctan (\frac{2 x - 1}{3})

簡素化

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

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

定数を解答に追加する

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