integral of (x^3-2x+3)/(x^2-x+1)

解

\int \frac{x ^{3} - 2 x + 3}{x ^{2} - x + 1} d x

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

解答ステップ

\int \frac{x ^{3} - 2 x + 3}{x ^{2} - x + 1} d x

拡張

\frac{x ^{3} - 2 x + 3}{x ^{2} - x + 1} : \frac{x ^{3}}{x ^{2} - x + 1} - \frac{2 x}{x ^{2} - x + 1} + \frac{3}{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{x ^{3}}{x ^{2} - x + 1} d x - \int \frac{2 x}{x ^{2} - x + 1} d x + \int \frac{3}{x ^{2} - x + 1} d x

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

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

\int \frac{3}{x ^{2} - x + 1} d x = 23 arctan (\frac{1}{3} (2 x - 1))

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

簡素化

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

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

定数を解答に追加する

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