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人気のある微分積分 >

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

解

\int \frac{- x}{( 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}{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

= - \int \frac{x}{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}{( 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

integral of 1/(xsqrt(1-(ln(x^2))))

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

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

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

integral of (6x+4)/(3x^2+4x)

\int \frac{6 x + 4}{3 x ^{2} + 4 x} d x

integral of 1+\sqrt[3]{x}

\int 1 + 3 x d x

integral of csc(-3h)

\int csc (- 3 h) d h