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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

u置換積分法を適用する

= - \int ln (u^{2} + \frac{3}{4}) d u

部分積分を適用する

= - (u ln (u^{2} + \frac{3}{4}) - \int \frac{8 u ^{2}}{4 u ^{2} + 3} d u)

\int \frac{8 u ^{2}}{4 u ^{2} + 3} d u = - 3 arctan (\frac{2}{3} u) + 2 u

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

代用を戻す

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

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

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

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\int \frac{1}{x ^{\frac{1}{n}}} d x

integral of (sin(11x))/(1+cos^2(11x))

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