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

解

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

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

解答ステップ

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

拡張

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

\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))

= - (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

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

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

定数を解答に追加する

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