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

解

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

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

解答ステップ

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

拡張

\frac{1 + x ^{2}}{1 + x ^{4}} : \frac{1}{1 + x ^{4}} + \frac{x ^{2}}{1 + x ^{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{1}{1 + x ^{4}} d x + \int \frac{x ^{2}}{1 + x ^{4}} d x

\int \frac{1}{1 + x ^{4}} d x = \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1)) + \frac{1}{4 2} (- ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1))

\int \frac{x ^{2}}{1 + x ^{4}} d x = - \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)) + \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1))

= \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1)) + \frac{1}{4 2} (- ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)) - \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)) + \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1))

簡素化

\frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1)) + \frac{1}{4 2} (- ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)) - \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)) + \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)) : \frac{1}{2} arctan (2 x + 1) + \frac{1}{2} arctan (2 x - 1)

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

定数を解答に追加する

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