integral from 0 to 3 of arctan(sqrt(x))

解答

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

\frac{4 π - 3 3}{3}

求解步骤

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

使用换元积分法

: \int_{0}^{3} arctan (u) \cdot 2 u d u

= \int_{0}^{3} arctan (u) \cdot 2 u d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{3} arctan (u) u d u

使用分布积分法

: [\frac{1}{2} (u^{2} arctan (u) - 2 \cdot \int \frac{u ^{2}}{2 ( u ^{2} + 1 )} d u)]_{0}^{3}

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

\int \frac{u ^{2}}{2 ( u ^{2} + 1 )} d u = \frac{1}{2} (- arctan (u) + u)

= 2 [\frac{1}{2} (u^{2} arctan (u) - 2 \cdot \frac{1}{2} (- arctan (u) + u))]_{0}^{3}

化简

2 [\frac{1}{2} (u^{2} arctan (u) - 2 \cdot \frac{1}{2} (- arctan (u) + u))]_{0}^{3} : 2 [\frac{1}{2} (u^{2} arctan (u) + arctan (u) - u)]_{0}^{3}

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

带入上下限计算后得:

\frac{4 π - 3 3}{6}

= 2 \cdot \frac{4 π - 3 3}{6}

化简

= \frac{4 π - 3 3}{3}

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