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人気のある微分積分 >

integral from e to 3 of 1/(16-x^2)

解

\int_{e}^{3} \frac{1}{16 - x ^{2}} d x

解

\frac{ln ( 7 ) - ln ( \frac{e}{4} + 1 ) + ln ( - \frac{e}{4} + 1 )}{8}

+1

十進法表記

0.03615 \dots

解答ステップ

\int_{e}^{3} \frac{1}{16 - x ^{2}} d x

置換積分法を適用する

= \int_{\frac{e}{4}}^{\frac{3}{4}} \frac{1}{4 ( - 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}{4} \cdot \int_{\frac{e}{4}}^{\frac{3}{4}} \frac{1}{- u ^{2} + 1} d u

共通積分を使用する:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{4} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{e}{4}}^{\frac{3}{4}}

簡素化

\frac{1}{4} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{e}{4}}^{\frac{3}{4}} : \frac{1}{4} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{e}{4}}^{\frac{3}{4}}

= \frac{1}{4} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{e}{4}}^{\frac{3}{4}}

限度を計算する:

\frac{ln ( \frac{7}{4} ) + 2 ln ( 2 ) - ln ( \frac{e}{4} + 1 ) + ln ( - \frac{e}{4} + 1 )}{2}

= \frac{1}{4} \cdot \frac{ln ( \frac{7}{4} ) + 2 ln ( 2 ) - ln ( \frac{e}{4} + 1 ) + ln ( - \frac{e}{4} + 1 )}{2}

簡素化

= \frac{ln ( 7 ) - ln ( \frac{e}{4} + 1 ) + ln ( - \frac{e}{4} + 1 )}{8}

グラフ

人気の例

integral from 0 to pi of 1+sin(x)

\int_{0}^{π} 1 + sin (x) d x

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(\partial)/(\partial y)(cos(x)+sin(y))

\frac{\partial}{\partial y} (cos (x) + sin (y))