integral from 0 to 1 of arctan(x^2)

解

\int_{0}^{1} arctan (x^{2}) d x

\frac{π - 2 ( 2 arctan ( 1 + 2 ) + 2 arctan ( 2 - 1 ) + ln ( 3 - 2 2 ) )}{4}

十進法表記

0.29790 \dots

解答ステップ

\int_{0}^{1} arctan (x^{2}) d x

部分積分を適用する

= [x arctan (x^{2}) - \int \frac{2 x ^{2}}{x ^{4} + 1} d x]_{0}^{1}

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

= [x arctan (x^{2}) - 2 (- \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)))]_{0}^{1}

限度を計算する:

\frac{π - 2 ( 2 arctan ( 1 + 2 ) + 2 arctan ( 2 - 1 ) + ln ( 3 - 2 2 ) )}{4}

= \frac{π - 2 ( 2 arctan ( 1 + 2 ) + 2 arctan ( 2 - 1 ) + ln ( 3 - 2 2 ) )}{4}