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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

簡素化

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

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

定数を解答に追加する

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