integral of (4x^2-7x-19)/(x^2-2x-3)

解

\int \frac{4 x ^{2} - 7 x - 19}{x ^{2} - 2 x - 3} d x

4 x + 2 ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ - 4 + C

解答ステップ

\int \frac{4 x ^{2} - 7 x - 19}{x ^{2} - 2 x - 3} d x

拡張

\frac{4 x ^{2} - 7 x - 19}{x ^{2} - 2 x - 3} : \frac{4 x ^{2}}{x ^{2} - 2 x - 3} - \frac{7 x}{x ^{2} - 2 x - 3} - \frac{19}{x ^{2} - 2 x - 3}

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{4 x ^{2}}{x ^{2} - 2 x - 3} d x - \int \frac{7 x}{x ^{2} - 2 x - 3} d x - \int \frac{19}{x ^{2} - 2 x - 3} d x

\int \frac{4 x ^{2}}{x ^{2} - 2 x - 3} d x = 4 x - 4 - ln ∣ x + 1 ∣ + 9 ln ∣ x - 3 ∣

\int \frac{7 x}{x ^{2} - 2 x - 3} d x = 7 (\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣)

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

= 4 x - 4 - ln ∣ x + 1 ∣ + 9 ln ∣ x - 3 ∣ - 7 (\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣) - 19 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣)

簡素化

4 x - 4 - ln ∣ x + 1 ∣ + 9 ln ∣ x - 3 ∣ - 7 (\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣) - 19 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣) : 4 x + 2 ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ - 4

= 4 x + 2 ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ - 4

定数を解答に追加する

= 4 x + 2 ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ - 4 + C