integral from 0 to 1 of arctan(x^3)

解答

\int_{0}^{1} arctan (x^{3}) d x

\frac{3 π - 2 π + 2 3 ln ( 2 )}{4 3}

十进制

0.22507 \dots

求解步骤

\int_{0}^{1} arctan (x^{3}) d x

使用分布积分法

= [x arctan (x^{3}) - \int \frac{3 x ^{3}}{x ^{6} + 1} d x]_{0}^{1}

\int \frac{3 x ^{3}}{x ^{6} + 1} d x = \frac{1}{4} (- 2 ln x^{2} + 1 + ln 4 x^{4} - 4 x^{2} + 4 + 23 arctan (\frac{2 x ^{2} - 1}{3}))

= [x arctan (x^{3}) - \frac{1}{4} (- 2 ln x^{2} + 1 + ln 4 x^{4} - 4 x^{2} + 4 + 23 arctan (\frac{2 x ^{2} - 1}{3}))]_{0}^{1}

带入上下限计算后得:

\frac{3 π - 2 π + 2 3 ln ( 2 )}{4 3}

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

\int_{0}^{e} x ln (2 x) d x

\int_{l n (2)}^{l n (10)} tanh (x) d x

\int_{0}^{x^{2}} \frac{4 t - 1}{t ^{2} + 1} d t

\int_{0}^{t} cos^{2} (2 π x) d x

\int_{1}^{3} (4^{x + 1} + 2^{x}) d x