integral of (8-3x)/(x^2+x+1)

解

\int \frac{8 - 3 x}{x ^{2} + x + 1} d x

\frac{16}{3} arctan (\frac{2 x + 1}{3}) - 6 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) + C

解答ステップ

\int \frac{8 - 3 x}{x ^{2} + x + 1} d x

拡張

\frac{8 - 3 x}{x ^{2} + x + 1} : \frac{8}{x ^{2} + x + 1} - \frac{3 x}{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{8}{x ^{2} + x + 1} d x - \int \frac{3 x}{x ^{2} + x + 1} d x

\int \frac{8}{x ^{2} + x + 1} d x = \frac{16}{3} arctan (\frac{2 x + 1}{3})

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

= \frac{16}{3} arctan (\frac{2 x + 1}{3}) - 6 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3}))

定数を解答に追加する

= \frac{16}{3} arctan (\frac{2 x + 1}{3}) - 6 (\frac{1}{4} ln 4 x^{2} + 4 x + 4 - \frac{1}{2 3} arctan (\frac{2 x + 1}{3})) + C