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人気のある微分積分 >

integral from-2 to 0 of 2/(x^2+4x+8)

解

\int_{- 2}^{0} \frac{2}{x ^{2} + 4 x + 8} d x

解

\frac{π}{4}

+1

十進法表記

0.78539 \dots

解答ステップ

\int_{- 2}^{0} \frac{2}{x ^{2} + 4 x + 8} d x

定数を除く:

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

= 2 \cdot \int_{- 2}^{0} \frac{1}{x ^{2} + 4 x + 8} d x

平方完成する

x^{2} + 4 x + 8 : (x + 2)^{2} + 4

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

u置換積分法を適用する

= 2 \cdot \int_{0}^{2} \frac{1}{u ^{2} + 4} d u

置換積分法を適用する

= 2 \cdot \int_{0}^{1} \frac{1}{2 ( v ^{2} + 1 )} d v

定数を除く:

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

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{1} \frac{1}{v ^{2} + 1} d v

共通積分を使用する:

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

= 2 \cdot \frac{1}{2} [arctan (v)]_{0}^{1}

簡素化

2 \cdot \frac{1}{2} [arctan (v)]_{0}^{1} : [arctan (v)]_{0}^{1}

= [arctan (v)]_{0}^{1}

限度を計算する:

\frac{π}{4}

= \frac{π}{4}

グラフ

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