integral of (2x^5+1)/(2x-2x^5)

解

\int \frac{2 x ^{5} + 1}{2 x - 2 x ^{5}} d x

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

解答ステップ

\int \frac{2 x ^{5} + 1}{2 x - 2 x ^{5}} d x

以下の部分分数を得る:

\frac{2 x ^{5} + 1}{2 x - 2 x ^{5}} : - 1 + \frac{1}{2 x} + \frac{- x + 2}{4 ( x ^{2} + 1 )} + \frac{1}{8 ( x + 1 )} - \frac{3}{8 ( x - 1 )}

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

\int 1 d x = x

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

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

\int \frac{1}{8 ( x + 1 )} d x = \frac{1}{8} ln ∣ x + 1 ∣

\int \frac{3}{8 ( x - 1 )} d x = \frac{3}{8} ln ∣ x - 1 ∣

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

定数を解答に追加する

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