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

解

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

\frac{1}{2} x - \frac{1}{12} ln ∣ 2 x - 1 ∣ - \frac{1}{3} ln x^{2} - x + 1 + C

解答ステップ

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

以下の部分分数を得る:

\frac{( x - 1 ) ^{3}}{x ^{3} + ( x - 1 ) ^{3}} : \frac{1}{2} - \frac{1}{6 ( 2 x - 1 )} + \frac{- 2 x + 1}{3 ( x ^{2} - x + 1 )}

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

\int \frac{1}{2} d x = \frac{1}{2} x

\int \frac{1}{6 ( 2 x - 1 )} d x = \frac{1}{12} ln ∣ 2 x - 1 ∣

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

= \frac{1}{2} x - \frac{1}{12} ln ∣ 2 x - 1 ∣ - \frac{1}{3} ln x^{2} - x + 1

定数を解答に追加する

= \frac{1}{2} x - \frac{1}{12} ln ∣ 2 x - 1 ∣ - \frac{1}{3} ln x^{2} - x + 1 + C