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人気のある微分積分 >

integral from 1 to 3 of 1/(x^2+4)

解

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

解

\frac{1}{2} (arctan (\frac{3}{2}) - arctan (\frac{1}{2}))

+1

十進法表記

0.25957 \dots

解答ステップ

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

置換積分法を適用する

= \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{2 ( u ^{2} + 1 )} d u

定数を除く:

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

= \frac{1}{2} \cdot \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{1}{u ^{2} + 1} d u

共通積分を使用する:

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

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

限度を計算する:

arctan (\frac{3}{2}) - arctan (\frac{1}{2})

= \frac{1}{2} (arctan (\frac{3}{2}) - arctan (\frac{1}{2}))

グラフ

人気の例

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\int_{\frac{1}{3}}^{3} (2 - \frac{2}{x ^{2}}) d x

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\int_{0}^{4} 1000 x d x

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\int_{0}^{π} cos (x) sin^{3} (x) d x

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