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人気のある微分積分 >

integral of (-x+1)/(3(x^2+x+1))

解

\int \frac{- x + 1}{3 ( x ^{2} + x + 1 )} d x

解

\frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} + 4 x}{3} + 23 arctan (\frac{2 x + 1}{3})) + C

解答ステップ

\int \frac{- x + 1}{3 ( x ^{2} + x + 1 )} d x

定数を除く:

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

= \frac{1}{3} \cdot \int \frac{- x + 1}{x ^{2} + x + 1} d x

平方完成する

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

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

u置換積分法を適用する

= \frac{1}{3} \cdot \int \frac{2 ( - 2 u + 3 )}{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

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

置換積分法を適用する

= \frac{1}{3} \cdot 2 \cdot \int \frac{- v + 3}{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

= \frac{1}{3} \cdot 2 \cdot \frac{1}{2} \cdot \int \frac{- v + 3}{v ^{2} + 1} d v

拡張

\frac{- v + 3}{v ^{2} + 1} : - \frac{v}{v ^{2} + 1} + \frac{3}{v ^{2} + 1}

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

= \frac{1}{3} \cdot 2 \cdot \frac{1}{2} (- \int \frac{v}{v ^{2} + 1} d v + \int \frac{3}{v ^{2} + 1} d v)

\int \frac{v}{v ^{2} + 1} d v = \frac{1}{2} ln v^{2} + 1

\int \frac{3}{v ^{2} + 1} d v = 3 arctan (v)

= \frac{1}{3} \cdot 2 \cdot \frac{1}{2} (- \frac{1}{2} ln v^{2} + 1 + 3 arctan (v))

代用を戻す

= \frac{1}{3} \cdot 2 \cdot \frac{1}{2} (- \frac{1}{2} ln (\frac{2}{3} (x + \frac{1}{2}))^{2} + 1 + 3 arctan (\frac{2}{3} (x + \frac{1}{2})))

簡素化

\frac{1}{3} \cdot 2 \cdot \frac{1}{2} (- \frac{1}{2} ln (\frac{2}{3} (x + \frac{1}{2}))^{2} + 1 + 3 arctan (\frac{2}{3} (x + \frac{1}{2}))) : \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} + 4 x}{3} + 23 arctan (\frac{2 x + 1}{3}))

= \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} + 4 x}{3} + 23 arctan (\frac{2 x + 1}{3}))

定数を解答に追加する

= \frac{1}{6} (- ln \frac{4}{3} + \frac{4 x ^{2} + 4 x}{3} + 23 arctan (\frac{2 x + 1}{3})) + C

グラフ

人気の例

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