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人気のある微分積分 >

integral from 0 to pi of 1/(4x^2-4x+5)

解

\int_{0}^{π} \frac{1}{4 x ^{2} - 4 x + 5} d x

解

\frac{1}{4} (arctan (π - \frac{1}{2}) + arctan (\frac{1}{2}))

+1

十進法表記

0.41813 \dots

解答ステップ

\int_{0}^{π} \frac{1}{4 x ^{2} - 4 x + 5} d x

平方完成する

4 x^{2} - 4 x + 5 : 4 (x - \frac{1}{2})^{2} + 4

= \int_{0}^{π} \frac{1}{4 ( x - \frac{1}{2} ) ^{2} + 4} d x

u置換積分法を適用する

= \int_{- \frac{1}{2}}^{π - \frac{1}{2}} \frac{1}{4 u ^{2} + 4} d u

\frac{1}{4 u ^{2} + 4} = \frac{1}{4} \frac{1}{u ^{2} + 1}

= \int_{- \frac{1}{2}}^{π - \frac{1}{2}} \frac{1}{4} \cdot \frac{1}{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{1}{2}}^{π - \frac{1}{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}{4} [arctan (u)]_{- \frac{1}{2}}^{π - \frac{1}{2}}

限度を計算する:

arctan (π - \frac{1}{2}) + arctan (\frac{1}{2})

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

グラフ

人気の例

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