integral of ((11x-7))/(2x^2-3x-2)

解

\int \frac{( 11 x - 7 )}{2 x ^{2} - 3 x - 2} d x

\frac{5}{2} ln ∣ 2 x + 1 ∣ + 3 ln ∣ x - 2 ∣ + C

解答ステップ

\int \frac{11 x - 7}{2 x ^{2} - 3 x - 2} d x

拡張

\frac{11 x - 7}{2 x ^{2} - 3 x - 2} : \frac{11 x}{2 x ^{2} - 3 x - 2} - \frac{7}{2 x ^{2} - 3 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{11 x}{2 x ^{2} - 3 x - 2} d x - \int \frac{7}{2 x ^{2} - 3 x - 2} d x

\int \frac{11 x}{2 x ^{2} - 3 x - 2} d x = 11 (\frac{1}{10} ln ∣ 2 x + 1 ∣ + \frac{2}{5} ln ∣ x - 2 ∣)

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

= 11 (\frac{1}{10} ln ∣ 2 x + 1 ∣ + \frac{2}{5} ln ∣ x - 2 ∣) - 7 (- \frac{1}{5} ln ∣ 2 x + 1 ∣ + \frac{1}{5} ln ∣ x - 2 ∣)

簡素化

11 (\frac{1}{10} ln ∣ 2 x + 1 ∣ + \frac{2}{5} ln ∣ x - 2 ∣) - 7 (- \frac{1}{5} ln ∣ 2 x + 1 ∣ + \frac{1}{5} ln ∣ x - 2 ∣) : \frac{5}{2} ln ∣ 2 x + 1 ∣ + 3 ln ∣ x - 2 ∣

= \frac{5}{2} ln ∣ 2 x + 1 ∣ + 3 ln ∣ x - 2 ∣

定数を解答に追加する

= \frac{5}{2} ln ∣ 2 x + 1 ∣ + 3 ln ∣ x - 2 ∣ + C