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人気のある微分積分 >

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

解

\int_{- 1}^{1} \frac{x}{x ^{2} + x + 1} d x

解

2 (\frac{1}{4} ln (3) - \frac{π}{4 3})

+1

十進法表記

- 0.35759 \dots

解答ステップ

\int_{- 1}^{1} \frac{x}{x ^{2} + x + 1} d x

平方完成する

x^{2} + x + 1 : (x + \frac{1}{2})^{2} + \frac{3}{4}

= \int_{- 1}^{1} \frac{x}{( x + \frac{1}{2} ) ^{2} + \frac{3}{4}} d x

u置換積分法を適用する

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

定数を除く:

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

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

拡張

\frac{2 u - 1}{4 u ^{2} + 3} : \frac{2 u}{4 u ^{2} + 3} - \frac{1}{4 u ^{2} + 3}

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int_{- \frac{1}{2}}^{\frac{3}{2}} \frac{1}{4 u ^{2} + 3} d u = \frac{π}{4 3}

= 2 (\frac{1}{4} ln (3) - \frac{π}{4 3})

グラフ

人気の例

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