integral of (x^3+2x^2+3x+7)/(x^2-x-6)

解

\int \frac{x ^{3} + 2 x ^{2} + 3 x + 7}{x ^{2} - x - 6} d x

\frac{x ^{2}}{2} + 3 x - \frac{8}{5} ln ∣ 2 x + 4 ∣ - \frac{13}{5} ln ∣ x + 2 ∣ + \frac{13}{5} ln ∣ x - 3 ∣ + \frac{18}{5} ln ∣ 2 x - 6 ∣ + 5 ln 4 x^{2} - 4 x - 24 - ln \frac{2}{5} x + \frac{4}{5} + ln \frac{2}{5} x - \frac{6}{5} - 1 + C

解答ステップ

\int \frac{x ^{3} + 2 x ^{2} + 3 x + 7}{x ^{2} - x - 6} d x

拡張

\frac{x ^{3} + 2 x ^{2} + 3 x + 7}{x ^{2} - x - 6} : \frac{x ^{3}}{x ^{2} - x - 6} + \frac{2 x ^{2}}{x ^{2} - x - 6} + \frac{3 x}{x ^{2} - x - 6} + \frac{7}{x ^{2} - x - 6}

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

= \int \frac{x ^{3}}{x ^{2} - x - 6} d x + \int \frac{2 x ^{2}}{x ^{2} - x - 6} d x + \int \frac{3 x}{x ^{2} - x - 6} d x + \int \frac{7}{x ^{2} - x - 6} d x

\int \frac{x ^{3}}{x ^{2} - x - 6} d x = \frac{x ^{2}}{2} + x + 14 (\frac{1}{4} ln 4 x^{2} - 4 x - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} x - ln - \frac{6}{5} + \frac{2}{5} x)) + 6 (- \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣)

\int \frac{2 x ^{2}}{x ^{2} - x - 6} d x = 2 x - 1 - \frac{8}{5} ln ∣ 2 x + 4 ∣ + \frac{18}{5} ln ∣ 2 x - 6 ∣

\int \frac{3 x}{x ^{2} - x - 6} d x = 6 (\frac{1}{4} ln 4 x^{2} - 4 x - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} x - ln - \frac{6}{5} + \frac{2}{5} x))

\int \frac{7}{x ^{2} - x - 6} d x = 7 (- \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣)

= \frac{x ^{2}}{2} + x + 14 (\frac{1}{4} ln 4 x^{2} - 4 x - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} x - ln - \frac{6}{5} + \frac{2}{5} x)) + 6 (- \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣) + 2 x - 1 - \frac{8}{5} ln ∣ 2 x + 4 ∣ + \frac{18}{5} ln ∣ 2 x - 6 ∣ + 6 (\frac{1}{4} ln 4 x^{2} - 4 x - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} x - ln - \frac{6}{5} + \frac{2}{5} x)) + 7 (- \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣)

簡素化

\frac{x ^{2}}{2} + x + 14 (\frac{1}{4} ln 4 x^{2} - 4 x - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} x - ln - \frac{6}{5} + \frac{2}{5} x)) + 6 (- \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣) + 2 x - 1 - \frac{8}{5} ln ∣ 2 x + 4 ∣ + \frac{18}{5} ln ∣ 2 x - 6 ∣ + 6 (\frac{1}{4} ln 4 x^{2} - 4 x - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} x - ln - \frac{6}{5} + \frac{2}{5} x)) + 7 (- \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣) : \frac{x ^{2}}{2} + 3 x - \frac{8}{5} ln ∣ 2 x + 4 ∣ - \frac{13}{5} ln ∣ x + 2 ∣ + \frac{13}{5} ln ∣ x - 3 ∣ + \frac{18}{5} ln ∣ 2 x - 6 ∣ + 5 ln 4 x^{2} - 4 x - 24 - ln \frac{2}{5} x + \frac{4}{5} + ln \frac{2}{5} x - \frac{6}{5} - 1

= \frac{x ^{2}}{2} + 3 x - \frac{8}{5} ln ∣ 2 x + 4 ∣ - \frac{13}{5} ln ∣ x + 2 ∣ + \frac{13}{5} ln ∣ x - 3 ∣ + \frac{18}{5} ln ∣ 2 x - 6 ∣ + 5 ln 4 x^{2} - 4 x - 24 - ln \frac{2}{5} x + \frac{4}{5} + ln \frac{2}{5} x - \frac{6}{5} - 1

定数を解答に追加する

= \frac{x ^{2}}{2} + 3 x - \frac{8}{5} ln ∣ 2 x + 4 ∣ - \frac{13}{5} ln ∣ x + 2 ∣ + \frac{13}{5} ln ∣ x - 3 ∣ + \frac{18}{5} ln ∣ 2 x - 6 ∣ + 5 ln 4 x^{2} - 4 x - 24 - ln \frac{2}{5} x + \frac{4}{5} + ln \frac{2}{5} x - \frac{6}{5} - 1 + C