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人気のある微分積分 >

integral of x/(x^2-x+6)

解

\int \frac{x}{x ^{2} - x + 6} d x

解

\frac{1}{2} ln 4 x^{2} - 4 x + 24 + \frac{1}{23} arctan (\frac{2 x - 1}{23}) + C

解答ステップ

\int \frac{x}{x ^{2} - x + 6} d x

平方完成する

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

= \int \frac{x}{( x - \frac{1}{2} ) ^{2} + \frac{23}{4}} d x

u置換積分法を適用する

= \int \frac{2 ( 2 u + 1 )}{4 u ^{2} + 23} 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{2 u + 1}{4 u ^{2} + 23} d u

拡張

\frac{2 u + 1}{4 u ^{2} + 23} : \frac{2 u}{4 u ^{2} + 23} + \frac{1}{4 u ^{2} + 23}

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{2 u}{4 u ^{2} + 23} d u + \int \frac{1}{4 u ^{2} + 23} d u)

\int \frac{2 u}{4 u ^{2} + 23} d u = \frac{1}{4} ln 4 u^{2} + 23

\int \frac{1}{4 u ^{2} + 23} d u = \frac{1}{2 23} arctan (\frac{2}{23} u)

= 2 (\frac{1}{4} ln 4 u^{2} + 23 + \frac{1}{2 23} arctan (\frac{2}{23} u))

代用を戻す

u = x - \frac{1}{2}

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

簡素化

2 (\frac{1}{4} ln 4 (x - \frac{1}{2})^{2} + 23 + \frac{1}{2 23} arctan (\frac{2}{23} (x - \frac{1}{2}))) : \frac{1}{2} ln 4 x^{2} - 4 x + 24 + \frac{1}{23} arctan (\frac{2 x - 1}{23})

= \frac{1}{2} ln 4 x^{2} - 4 x + 24 + \frac{1}{23} arctan (\frac{2 x - 1}{23})

定数を解答に追加する

= \frac{1}{2} ln 4 x^{2} - 4 x + 24 + \frac{1}{23} arctan (\frac{2 x - 1}{23}) + C

グラフ

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