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

解

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

8 (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1))))) - \frac{2}{3 3} (2 arctan (\frac{2 x + 1}{3}) + sin (2 arctan (\frac{2 x + 1}{3}))) + C

解答ステップ

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

拡張

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

\int \frac{x}{( x ^{2} + x + 1 ) ^{2}} d x = 8 (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1)))))

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

= 8 (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1))))) - \frac{2}{3 3} (2 arctan (\frac{2 x + 1}{3}) + sin (2 arctan (\frac{2 x + 1}{3})))

定数を解答に追加する

= 8 (- \frac{1}{4 ( 4 x ^{2} + 4 x + 4 )} - \frac{1}{24 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1))))) - \frac{2}{3 3} (2 arctan (\frac{2 x + 1}{3}) + sin (2 arctan (\frac{2 x + 1}{3}))) + C