integral from 0 to 0.1 of arctan(x^2)

解

\int_{0}^{0.1} arctan (x^{2}) d x

0.00033 \dots

解答ステップ

\int_{0}^{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}^{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}^{0.1}

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))) = 2 (- 0.17677 \dots (ln 2 x^{2} + 2.82842 \dots x + 2 - 2 arctan (1.41421 \dots x + 1)) + 0.17677 \dots (ln 2 x^{2} - 2.82842 \dots x + 2 + 2 arctan (1.41421 \dots x - 1)))

= [x arctan (x^{2}) - 2 (- 0.17677 \dots (ln 2 x^{2} + 2.82842 \dots x + 2 - 2 arctan (1.41421 \dots x + 1)) + 0.17677 \dots (ln 2 x^{2} - 2.82842 \dots x + 2 + 2 arctan (1.41421 \dots x - 1)))]_{0}^{0.1}

限度を計算する:

0.00033 \dots

= 0.00033 \dots