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人気のある微分積分 >

integral from 1 to 3 of arctan(sqrt(x))

解

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

解

\frac{5 π - 6 3 + 6}{6}

+1

十進法表記

1.88594 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{1}^{3} arctan (u) \cdot 2 u d u

定数を除く:

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

= 2 \cdot \int_{1}^{3} arctan (u) u d u

部分積分を適用する

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

簡素化

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

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

限度を計算する:

\frac{5 π - 6 3 + 6}{12}

= 2 \cdot \frac{5 π - 6 3 + 6}{12}

簡素化

= \frac{5 π - 6 3 + 6}{6}

グラフ

人気の例

integral from 0 to 4 of (8x+sqrt(x))

\int_{0}^{4} (8 x + x) d x

integral from-infinity to 0 of 1

\int_{- \infty}^{0} 1 d x

integral from 0 to 2 of sqrt(1+2x^2)

\int_{0}^{2} 1 + 2 x^{2} d x

integral from 0 to 2 of (-t^2+3t-2)

\int_{0}^{2} (- t^{2} + 3 t - 2) d t

integral from 0 to 4 of (8t-1)^2

\int_{0}^{4} (8 t - 1)^{2} d t