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人気のある微分積分 >

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

解

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

解

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

解答ステップ

\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}

= \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

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

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

\int \frac{1}{4 u ^{2} + 3} d u = \frac{1}{2 3} arctan (\frac{2}{3} u)

= 2 (\frac{1}{4} ln 4 u^{2} + 3 + \frac{1}{2 3} arctan (\frac{2}{3} u))

代用を戻す

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

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

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