integral of (x^2+3x+1)/(x^2-2x-3)

解

\int \frac{x ^{2} + 3 x + 1}{x ^{2} - 2 x - 3} d x

\frac{1}{4} ln ∣ x + 1 ∣ + \frac{19}{4} ln ∣ x - 3 ∣ + x + C

解答ステップ

\int \frac{x ^{2} + 3 x + 1}{x ^{2} - 2 x - 3} d x

\frac{x ^{2} + 3 x + 1}{x ^{2} - 2 x - 3} = \frac{5 x + 4}{x ^{2} - 2 x - 3} + 1

= \int \frac{5 x + 4}{x ^{2} - 2 x - 3} + 1 d x

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 + 4}{x ^{2} - 2 x - 3} d x + \int 1 d x

\int \frac{5 x + 4}{x ^{2} - 2 x - 3} d x = \frac{1}{4} ln ∣ x + 1 ∣ + \frac{19}{4} ln ∣ x - 3 ∣

\int 1 d x = x

= \frac{1}{4} ln ∣ x + 1 ∣ + \frac{19}{4} ln ∣ x - 3 ∣ + x

定数を解答に追加する

= \frac{1}{4} ln ∣ x + 1 ∣ + \frac{19}{4} ln ∣ x - 3 ∣ + x + C