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

解

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

- \frac{2}{9} ln ∣ x - 1 ∣ + \frac{1}{9} ln 4 x^{2} + 4 x + 4 + \frac{2}{9 3} arctan (\frac{1}{3} (2 x + 1)) + \frac{2}{3 3} arctan (\frac{1}{3} (2 x + 1)) + \frac{2}{9 3} sin (2 arctan (\frac{1}{3} (2 x + 1))) - \frac{1}{9 ( x - 1 )} - \frac{1}{6 ( x ^{2} + x + 1 )} - \frac{1}{9 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1)))) + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{( x ^{3} - 1 ) ^{2}} : - \frac{2}{9 ( x - 1 )} + \frac{1}{9 ( x - 1 ) ^{2}} + \frac{2 x + 3}{9 ( x ^{2} + x + 1 )} + \frac{x + 1}{3 ( x ^{2} + x + 1 ) ^{2}}

= \int - \frac{2}{9 ( x - 1 )} + \frac{1}{9 ( x - 1 ) ^{2}} + \frac{2 x + 3}{9 ( x ^{2} + x + 1 )} + \frac{x + 1}{3 ( x ^{2} + x + 1 ) ^{2}} d x

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}{9 ( x - 1 )} d x + \int \frac{1}{9 ( x - 1 ) ^{2}} d x + \int \frac{2 x + 3}{9 ( x ^{2} + x + 1 )} d x + \int \frac{x + 1}{3 ( x ^{2} + x + 1 ) ^{2}} d x

\int \frac{2}{9 ( x - 1 )} d x = \frac{2}{9} ln ∣ x - 1 ∣

\int \frac{1}{9 ( x - 1 ) ^{2}} d x = - \frac{1}{9 ( x - 1 )}

\int \frac{2 x + 3}{9 ( x ^{2} + x + 1 )} d x = \frac{1}{9} (ln 4 x^{2} + 4 x + 4 - \frac{2}{3} arctan (\frac{2 x + 1}{3}) + 23 arctan (\frac{1}{3} (2 x + 1)))

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

= - \frac{2}{9} ln ∣ x - 1 ∣ - \frac{1}{9 ( x - 1 )} + \frac{1}{9} (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{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}))))

簡素化

- \frac{2}{9} ln ∣ x - 1 ∣ - \frac{1}{9 ( x - 1 )} + \frac{1}{9} (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{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})))) : - \frac{2}{9} ln ∣ x - 1 ∣ + \frac{1}{9} ln 4 x^{2} + 4 x + 4 + \frac{2}{9 3} arctan (\frac{1}{3} (2 x + 1)) + \frac{2}{3 3} arctan (\frac{1}{3} (2 x + 1)) + \frac{2}{9 3} sin (2 arctan (\frac{1}{3} (2 x + 1))) - \frac{1}{9 ( x - 1 )} - \frac{1}{6 ( x ^{2} + x + 1 )} - \frac{1}{9 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1))))

= - \frac{2}{9} ln ∣ x - 1 ∣ + \frac{1}{9} ln 4 x^{2} + 4 x + 4 + \frac{2}{9 3} arctan (\frac{1}{3} (2 x + 1)) + \frac{2}{3 3} arctan (\frac{1}{3} (2 x + 1)) + \frac{2}{9 3} sin (2 arctan (\frac{1}{3} (2 x + 1))) - \frac{1}{9 ( x - 1 )} - \frac{1}{6 ( x ^{2} + x + 1 )} - \frac{1}{9 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1))))

定数を解答に追加する

= - \frac{2}{9} ln ∣ x - 1 ∣ + \frac{1}{9} ln 4 x^{2} + 4 x + 4 + \frac{2}{9 3} arctan (\frac{1}{3} (2 x + 1)) + \frac{2}{3 3} arctan (\frac{1}{3} (2 x + 1)) + \frac{2}{9 3} sin (2 arctan (\frac{1}{3} (2 x + 1))) - \frac{1}{9 ( x - 1 )} - \frac{1}{6 ( x ^{2} + x + 1 )} - \frac{1}{9 3} (2 arctan (\frac{1}{3} (2 x + 1)) + sin (2 arctan (\frac{1}{3} (2 x + 1)))) + C