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

解

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

- x^{2} arctan (x^{4}) + \frac{2 π - 4 arctan ( 1 + 2 ) - 4 arctan ( 2 - 1 ) - ln ( ( 17 - 12 2 ) 2 x ^{4} + 2 2 x ^{2} + 2 ^{2} ) + 4 arctan ( 2 x ^{2} + 1 ) + 2 ln 2 x ^{4} - 2 2 x ^{2} + 2 + 4 arctan ( 2 x ^{2} - 1 )}{4 2}

解答ステップ

\int_{x^{2}}^{1} 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]_{x^{2}}^{1}

\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)))]_{x^{2}}^{1}

限度を計算する:

\frac{π - 2 ( 2 arctan ( 1 + 2 ) + 2 arctan ( 2 - 1 ) + ln ( 3 - 2 2 ) )}{4} - x^{2} arctan (x^{4}) - \frac{1}{2 2} ln 2 x^{4} + 22 x^{2} + 2 + \frac{1}{2} arctan (2 x^{2} + 1) + \frac{1}{2 2} ln 2 x^{4} - 22 x^{2} + 2 + \frac{1}{2} arctan (2 x^{2} - 1)

= \frac{π - 2 ( 2 arctan ( 1 + 2 ) + 2 arctan ( 2 - 1 ) + ln ( 3 - 2 2 ) )}{4} - x^{2} arctan (x^{4}) - \frac{1}{2 2} ln 2 x^{4} + 22 x^{2} + 2 + \frac{1}{2} arctan (2 x^{2} + 1) + \frac{1}{2 2} ln 2 x^{4} - 22 x^{2} + 2 + \frac{1}{2} arctan (2 x^{2} - 1)

簡素化

= - x^{2} arctan (x^{4}) + \frac{2 π - 4 arctan ( 1 + 2 ) - 4 arctan ( 2 - 1 ) - ln ( ( 17 - 12 2 ) 2 x ^{4} + 2 2 x ^{2} + 2 ^{2} ) + 4 arctan ( 2 x ^{2} + 1 ) + 2 ln 2 x ^{4} - 2 2 x ^{2} + 2 + 4 arctan ( 2 x ^{2} - 1 )}{4 2}