integral of arctan(t^2)

解

\int arctan (t^{2}) d t

t arctan (t^{2}) - 2 (- \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 arctan (t^{2}) d t

部分積分を適用する

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

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

= t arctan (t^{2}) - 2 (- \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

= t arctan (t^{2}) - 2 (- \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