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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

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 8 sec^{3} (x) d x

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\int 0 cos (x) d x

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