integral of 1/(1+t^4)

解

\int \frac{1}{1 + t ^{4}} d t

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

解答ステップ

\int \frac{1}{1 + t ^{4}} d t

以下の部分分数を得る:

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

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

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

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

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

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

定数を解答に追加する

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