integral of 1/(9-x^4)

解

\int \frac{1}{9 - x ^{4}} d x

\frac{1}{6 3} arctan (\frac{x}{3}) + \frac{1}{12 3} ln x + 3 - \frac{1}{12 3} ln x - 3 + C

解答ステップ

\int \frac{1}{9 - x ^{4}} d x

以下の部分分数を得る:

\frac{1}{9 - x ^{4}} : \frac{1}{6 ( x ^{2} + 3 )} + \frac{1}{12 3 ( x + 3 )} - \frac{1}{12 3 ( x - 3 )}

= \int \frac{1}{6 ( x ^{2} + 3 )} + \frac{1}{12 3 ( x + 3 )} - \frac{1}{12 3 ( x - 3 )} 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}{6 ( x ^{2} + 3 )} d x + \int \frac{1}{12 3 ( x + 3 )} d x - \int \frac{1}{12 3 ( x - 3 )} d x

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

\int \frac{1}{12 3 ( x + 3 )} d x = \frac{1}{12 3} ln x + 3

\int \frac{1}{12 3 ( x - 3 )} d x = \frac{1}{12 3} ln x - 3

= \frac{1}{6 3} arctan (\frac{x}{3}) + \frac{1}{12 3} ln x + 3 - \frac{1}{12 3} ln x - 3

定数を解答に追加する

= \frac{1}{6 3} arctan (\frac{x}{3}) + \frac{1}{12 3} ln x + 3 - \frac{1}{12 3} ln x - 3 + C