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

解

\int \frac{t ^{2}}{t ^{4} + 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)) + C

解答ステップ

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

以下の部分分数を得る:

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

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

\int \frac{t}{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 ( 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