integral of (x+3)/(x^4+1)

解

\int \frac{x + 3}{x ^{4} + 1} d x

\frac{1}{2} arctan (x^{2}) + 3 (\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))) + C

解答ステップ

\int \frac{x + 3}{x ^{4} + 1} d x

拡張

\frac{x + 3}{x ^{4} + 1} : \frac{x}{x ^{4} + 1} + \frac{3}{x ^{4} + 1}

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}{x ^{4} + 1} d x + \int \frac{3}{x ^{4} + 1} d x

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

\int \frac{3}{x ^{4} + 1} d x = 3 (\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 (x^{2}) + 3 (\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)))

定数を解答に追加する

= \frac{1}{2} arctan (x^{2}) + 3 (\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))) + C