integral of 1/(x^2-x^4)

解

\int \frac{1}{x ^{2} - x ^{4}} d x

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

解答ステップ

\int \frac{1}{x ^{2} - x ^{4}} d x

以下の部分分数を得る:

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

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

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

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

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

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

定数を解答に追加する

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