integral of (5x-4)/(x^2+3x+2)

解

\int \frac{5 x - 4}{x ^{2} + 3 x + 2} d x

5 (- ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣) + 8 (\frac{1}{2} ln ∣ 2 x + 4 ∣ - \frac{1}{2} ln ∣ 2 x + 2 ∣) + C

解答ステップ

\int \frac{5 x - 4}{x ^{2} + 3 x + 2} d x

拡張

\frac{5 x - 4}{x ^{2} + 3 x + 2} : \frac{5 x}{x ^{2} + 3 x + 2} - \frac{4}{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{5 x}{x ^{2} + 3 x + 2} d x - \int \frac{4}{x ^{2} + 3 x + 2} d x

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

\int \frac{4}{x ^{2} + 3 x + 2} d x = - 8 (\frac{1}{2} ln ∣ 2 x + 4 ∣ - \frac{1}{2} ln ∣ 2 x + 2 ∣)

= 5 (- ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣) - (- 8 (\frac{1}{2} ln ∣ 2 x + 4 ∣ - \frac{1}{2} ln ∣ 2 x + 2 ∣))

簡素化

= 5 (- ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣) + 8 (\frac{1}{2} ln ∣ 2 x + 4 ∣ - \frac{1}{2} ln ∣ 2 x + 2 ∣)

定数を解答に追加する

= 5 (- ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣) + 8 (\frac{1}{2} ln ∣ 2 x + 4 ∣ - \frac{1}{2} ln ∣ 2 x + 2 ∣) + C