integral of (11x+2)/(2x^2-5x-3)

解

\int \frac{11 x + 2}{2 x ^{2} - 5 x - 3} d x

- \frac{1}{7} (11 x + 2) (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) + \frac{11}{14} (2 x (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) + ln ∣ 2 x + 1 ∣ + 6 ln ∣ 2 x - 6 ∣) + C

解答ステップ

\int \frac{11 x + 2}{2 x ^{2} - 5 x - 3} d x

部分積分を適用する

= - \frac{1}{7} (11 x + 2) (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) - \int - \frac{11}{7} (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) d x

\int - \frac{11}{7} (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) d x = - \frac{11}{14} (2 x (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) + ln ∣ 2 x + 1 ∣ + 6 ln ∣ 2 x - 6 ∣)

= - \frac{1}{7} (11 x + 2) (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) - (- \frac{11}{14} (2 x (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) + ln ∣ 2 x + 1 ∣ + 6 ln ∣ 2 x - 6 ∣))

簡素化

- \frac{1}{7} (11 x + 2) (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) - (- \frac{11}{14} (2 x (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) + ln ∣ 2 x + 1 ∣ + 6 ln ∣ 2 x - 6 ∣)) : - \frac{1}{7} (11 x + 2) (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) + \frac{11}{14} (2 x (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) + ln ∣ 2 x + 1 ∣ + 6 ln ∣ 2 x - 6 ∣)

= - \frac{1}{7} (11 x + 2) (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) + \frac{11}{14} (2 x (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) + ln ∣ 2 x + 1 ∣ + 6 ln ∣ 2 x - 6 ∣)

定数を解答に追加する

= - \frac{1}{7} (11 x + 2) (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) + \frac{11}{14} (2 x (ln \frac{2}{7} + \frac{4}{7} x - ln - \frac{12}{7} + \frac{4}{7} x) + ln ∣ 2 x + 1 ∣ + 6 ln ∣ 2 x - 6 ∣) + C