integral of (x^2)/(x^4-16)

解

\int \frac{x ^{2}}{x ^{4} - 16} d x

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

解答ステップ

\int \frac{x ^{2}}{x ^{4} - 16} d x

以下の部分分数を得る:

\frac{x ^{2}}{x ^{4} - 16} : \frac{1}{2 ( x ^{2} + 4 )} - \frac{1}{8 ( x + 2 )} + \frac{1}{8 ( x - 2 )}

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

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

\int \frac{1}{8 ( x + 2 )} d x = \frac{1}{8} ln ∣ x + 2 ∣

\int \frac{1}{8 ( x - 2 )} d x = \frac{1}{8} ln ∣ x - 2 ∣

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

定数を解答に追加する

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