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

解答

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

\frac{π}{12} + \frac{ln ( 2 ) - 1}{6}

十进制

0.21065 \dots

求解步骤

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

使用分布积分法

= [\frac{1}{3} x^{3} arctan (x) - \int \frac{x ^{3}}{3 ( x ^{2} + 1 )} d x]_{0}^{1}

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

= [\frac{1}{3} x^{3} arctan (x) - \frac{1}{6} (x^{2} + 1 - ln x^{2} + 1)]_{0}^{1}

带入上下限计算后得:

\frac{π}{12} - \frac{- ln ( 2 ) + 2}{6} + \frac{1}{6}

= \frac{π}{12} - \frac{- ln ( 2 ) + 2}{6} + \frac{1}{6}

化简

= \frac{π}{12} + \frac{ln ( 2 ) - 1}{6}

n = 1 \sum \infty \frac{10}{3 ^{n}}

\int_{0}^{1} \frac{1}{( x + 1 ) ( x - 2 )} d x

\int \frac{x ^{2} - a ^{2}}{x} d x

\frac{\partial}{\partial y} (cos (x^{2}) sin (yz))

\frac{d y}{d x} = \frac{y ^{2} + 4 x x ^{2} + y ^{2}}{x y}