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

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

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

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

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

定数を解答に追加する

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