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

解

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

6 (\frac{1}{8} ln 16 x^{2} - 40 x + 16 - \frac{5}{24} (ln - \frac{2}{3} + \frac{4}{3} x - ln - \frac{8}{3} + \frac{4}{3} x)) - 2 (- \frac{1}{3} ln ∣ 2 x - 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣) + C

解答ステップ

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

拡張

\frac{3 x - 2}{2 x ^{2} - 5 x + 2} : \frac{3 x}{2 x ^{2} - 5 x + 2} - \frac{2}{2 x ^{2} - 5 x + 2}

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

\int \frac{3 x}{2 x ^{2} - 5 x + 2} d x = 6 (\frac{1}{8} ln 16 x^{2} - 40 x + 16 - \frac{5}{24} (ln - \frac{2}{3} + \frac{4}{3} x - ln - \frac{8}{3} + \frac{4}{3} x))

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

= 6 (\frac{1}{8} ln 16 x^{2} - 40 x + 16 - \frac{5}{24} (ln - \frac{2}{3} + \frac{4}{3} x - ln - \frac{8}{3} + \frac{4}{3} x)) - 2 (- \frac{1}{3} ln ∣ 2 x - 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣)

定数を解答に追加する

= 6 (\frac{1}{8} ln 16 x^{2} - 40 x + 16 - \frac{5}{24} (ln - \frac{2}{3} + \frac{4}{3} x - ln - \frac{8}{3} + \frac{4}{3} x)) - 2 (- \frac{1}{3} ln ∣ 2 x - 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣) + C