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人気のある微分積分 >

integral of 7ln(x^2-x+8)

解

\int 7 ln (x^{2} - x + 8) d x

解

7 (x ln (x^{2} - x + 8) - \frac{1}{2} ln (x^{2} - x + 8) + 31 arctan (\frac{1}{31} (2 x - 1)) - 2 x + 1) + C

解答ステップ

\int 7 ln (x^{2} - x + 8) d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int ln (x^{2} - x + 8) d x

平方完成する

x^{2} - x + 8 : (x - \frac{1}{2})^{2} + \frac{31}{4}

= 7 \cdot \int ln ((x - \frac{1}{2})^{2} + \frac{31}{4}) d x

u置換積分法を適用する

= 7 \cdot \int ln (u^{2} + \frac{31}{4}) d u

部分積分を適用する

= 7 (u ln (u^{2} + \frac{31}{4}) - \int \frac{8 u ^{2}}{4 u ^{2} + 31} d u)

\int \frac{8 u ^{2}}{4 u ^{2} + 31} d u = - 31 arctan (\frac{2}{31} u) + 2 u

= 7 (u ln (u^{2} + \frac{31}{4}) - (- 31 arctan (\frac{2}{31} u) + 2 u))

代用を戻す

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

= 7 ((x - \frac{1}{2}) ln ((x - \frac{1}{2})^{2} + \frac{31}{4}) - (- 31 arctan (\frac{2}{31} (x - \frac{1}{2})) + 2 (x - \frac{1}{2})))

簡素化

7 ((x - \frac{1}{2}) ln ((x - \frac{1}{2})^{2} + \frac{31}{4}) - (- 31 arctan (\frac{2}{31} (x - \frac{1}{2})) + 2 (x - \frac{1}{2}))) : 7 (x ln (x^{2} - x + 8) - \frac{1}{2} ln (x^{2} - x + 8) + 31 arctan (\frac{1}{31} (2 x - 1)) - 2 x + 1)

= 7 (x ln (x^{2} - x + 8) - \frac{1}{2} ln (x^{2} - x + 8) + 31 arctan (\frac{1}{31} (2 x - 1)) - 2 x + 1)

定数を解答に追加する

= 7 (x ln (x^{2} - x + 8) - \frac{1}{2} ln (x^{2} - x + 8) + 31 arctan (\frac{1}{31} (2 x - 1)) - 2 x + 1) + C

グラフ

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