integral of ((x^2+2))/(x^4+4)

解

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

\frac{1}{2} arctan (x + 1) + \frac{1}{2} arctan (x - 1) + C

解答ステップ

\int \frac{x ^{2} + 2}{x ^{4} + 4} d x

拡張

\frac{x ^{2} + 2}{x ^{4} + 4} : \frac{x ^{2}}{x ^{4} + 4} + \frac{2}{x ^{4} + 4}

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{x ^{2}}{x ^{4} + 4} d x + \int \frac{2}{x ^{4} + 4} d x

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

\int \frac{2}{x ^{4} + 4} d x = 2 (\frac{1}{16} (ln x^{2} + 2 x + 2 + 2 arctan (x + 1)) + \frac{1}{16} (- ln x^{2} - 2 x + 2 + 2 arctan (x - 1)))

= - \frac{1}{8} (ln x^{2} + 2 x + 2 - 2 arctan (x + 1)) + \frac{1}{8} (ln x^{2} - 2 x + 2 + 2 arctan (x - 1)) + 2 (\frac{1}{16} (ln x^{2} + 2 x + 2 + 2 arctan (x + 1)) + \frac{1}{16} (- ln x^{2} - 2 x + 2 + 2 arctan (x - 1)))

簡素化

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

= \frac{1}{2} arctan (x + 1) + \frac{1}{2} arctan (x - 1)

定数を解答に追加する

= \frac{1}{2} arctan (x + 1) + \frac{1}{2} arctan (x - 1) + C