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人気のある微分積分 >

integral from 0 to x of 1/(t^2+t+1)

解

\int_{0}^{x} \frac{1}{t ^{2} + t + 1} d t

解

\frac{2}{3} arctan (\frac{2 x + 1}{3}) - \frac{π}{3 3}

解答ステップ

\int_{0}^{x} \frac{1}{t ^{2} + t + 1} d t

平方完成する

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

= \int_{0}^{x} \frac{1}{( t + \frac{1}{2} ) ^{2} + \frac{3}{4}} d t

u置換積分法を適用する

= \int_{\frac{1}{2}}^{x + \frac{1}{2}} \frac{4}{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

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

置換積分法を適用する

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

定数を除く:

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

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

= 4 \cdot \frac{1}{2 3} [arctan (v)]_{\frac{1}{3}}^{\frac{2}{3} (x + \frac{1}{2})}

簡素化

4 \cdot \frac{1}{2 3} [arctan (v)]_{\frac{1}{3}}^{\frac{2}{3} (x + \frac{1}{2})} : \frac{2}{3} [arctan (v)]_{\frac{1}{3}}^{\frac{2 x + 1}{3}}

= \frac{2}{3} [arctan (v)]_{\frac{1}{3}}^{\frac{2 x + 1}{3}}

限度を計算する:

arctan (\frac{2 x + 1}{3}) - \frac{π}{6}

= \frac{2}{3} (arctan (\frac{2 x + 1}{3}) - \frac{π}{6})

簡素化

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

人気の例

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\frac{d}{d x} (\frac{x + 7}{x ^{3} + x - 6})