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

解

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

- \frac{4}{3 3} arctan (\frac{1}{3} (2 x + 1)) + \frac{1}{3 3} sin (2 arctan (\frac{1}{3} (2 x + 1))) - \frac{1}{2 ( x ^{2} + x + 1 )} - \frac{8}{3 3} arctan (\frac{1}{3} (2 x + 1)) - \frac{4}{3 3} sin (2 arctan (\frac{1}{3} (2 x + 1))) + C

解答ステップ

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

拡張

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

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

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

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

簡素化

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

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

定数を解答に追加する

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