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人気のある微分積分 >

integral from 0 to 1 of 2piarctan(x)

解

\int_{0}^{1} 2 π arctan (x) d x

解

\frac{π ^{2}}{2} - π ln (2)

+1

十進法表記

2.75721 \dots

解答ステップ

\int_{0}^{1} 2 π arctan (x) d x

定数を除く:

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

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

部分積分を適用する

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

\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln x^{2} + 1

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

限度を計算する:

\frac{π}{4} - \frac{1}{2} ln (2)

= 2 π (\frac{π}{4} - \frac{1}{2} ln (2))

簡素化

= \frac{π ^{2}}{2} - π ln (2)

グラフ

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