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人気のある微分積分 >

integral of ln(x^2-x-6)

解

\int ln (x^{2} - x - 6) d x

解

x ln (x^{2} - x - 6) - \frac{1}{2} ln (x^{2} - x - 6) + \frac{5}{2} ln \frac{4}{5} + \frac{2}{5} x - \frac{5}{2} ln - \frac{6}{5} + \frac{2}{5} x - 2 x + 1 + C

解答ステップ

\int ln (x^{2} - x - 6) d x

平方完成する

x^{2} - x - 6 : (x - \frac{1}{2})^{2} - \frac{25}{4}

= \int ln ((x - \frac{1}{2})^{2} - \frac{25}{4}) d x

u置換積分法を適用する

= \int ln (u^{2} - \frac{25}{4}) d u

部分積分を適用する

= u ln (u^{2} - \frac{25}{4}) - \int \frac{8 u ^{2}}{4 u ^{2} - 25} d u

\int \frac{8 u ^{2}}{4 u ^{2} - 25} d u = - \frac{5}{2} ln \frac{2}{5} u + 1 + \frac{5}{2} ln \frac{2}{5} u - 1 + 2 u

= u ln (u^{2} - \frac{25}{4}) - (- \frac{5}{2} ln \frac{2}{5} u + 1 + \frac{5}{2} ln \frac{2}{5} u - 1 + 2 u)

代用を戻す

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

= (x - \frac{1}{2}) ln ((x - \frac{1}{2})^{2} - \frac{25}{4}) - (- \frac{5}{2} ln \frac{2}{5} (x - \frac{1}{2}) + 1 + \frac{5}{2} ln \frac{2}{5} (x - \frac{1}{2}) - 1 + 2 (x - \frac{1}{2}))

簡素化

(x - \frac{1}{2}) ln ((x - \frac{1}{2})^{2} - \frac{25}{4}) - (- \frac{5}{2} ln \frac{2}{5} (x - \frac{1}{2}) + 1 + \frac{5}{2} ln \frac{2}{5} (x - \frac{1}{2}) - 1 + 2 (x - \frac{1}{2})) : x ln (x^{2} - x - 6) - \frac{1}{2} ln (x^{2} - x - 6) + \frac{5}{2} ln \frac{4}{5} + \frac{2}{5} x - \frac{5}{2} ln - \frac{6}{5} + \frac{2}{5} x - 2 x + 1

= x ln (x^{2} - x - 6) - \frac{1}{2} ln (x^{2} - x - 6) + \frac{5}{2} ln \frac{4}{5} + \frac{2}{5} x - \frac{5}{2} ln - \frac{6}{5} + \frac{2}{5} x - 2 x + 1

定数を解答に追加する

= x ln (x^{2} - x - 6) - \frac{1}{2} ln (x^{2} - x - 6) + \frac{5}{2} ln \frac{4}{5} + \frac{2}{5} x - \frac{5}{2} ln - \frac{6}{5} + \frac{2}{5} x - 2 x + 1 + C

グラフ

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