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人気のある微分積分 >

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

解

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

解

\frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{1}{7} arctan (\frac{2 x + 1}{7}) + C

解答ステップ

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

平方完成する

x^{2} + x + 2 : (x + \frac{1}{2})^{2} + \frac{7}{4}

= \int \frac{x}{( x + \frac{1}{2} ) ^{2} + \frac{7}{4}} d x

u置換積分法を適用する

= \int \frac{2 ( 2 u - 1 )}{4 u ^{2} + 7} 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} + 7} d u

拡張

\frac{2 u - 1}{4 u ^{2} + 7} : \frac{2 u}{4 u ^{2} + 7} - \frac{1}{4 u ^{2} + 7}

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} + 7} d u - \int \frac{1}{4 u ^{2} + 7} d u)

\int \frac{2 u}{4 u ^{2} + 7} d u = \frac{1}{4} ln 4 u^{2} + 7

\int \frac{1}{4 u ^{2} + 7} d u = \frac{1}{2 7} arctan (\frac{2}{7} u)

= 2 (\frac{1}{4} ln 4 u^{2} + 7 - \frac{1}{2 7} arctan (\frac{2}{7} u))

代用を戻す

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

= 2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} + 7 - \frac{1}{2 7} arctan (\frac{2}{7} (x + \frac{1}{2})))

簡素化

2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} + 7 - \frac{1}{2 7} arctan (\frac{2}{7} (x + \frac{1}{2}))) : \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{1}{7} arctan (\frac{2 x + 1}{7})

= \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{1}{7} arctan (\frac{2 x + 1}{7})

定数を解答に追加する

= \frac{1}{2} ln 4 x^{2} + 4 x + 8 - \frac{1}{7} arctan (\frac{2 x + 1}{7}) + C

グラフ

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