integral of (4x^2-14x-6)/(x^3-2x^2-3x)

解

\int \frac{4 x ^{2} - 14 x - 6}{x ^{3} - 2 x ^{2} - 3 x} d x

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

解答ステップ

\int \frac{4 x ^{2} - 14 x - 6}{x ^{3} - 2 x ^{2} - 3 x} d x

拡張

\frac{4 x ^{2} - 14 x - 6}{x ^{3} - 2 x ^{2} - 3 x} : \frac{4 x ^{2}}{x ^{3} - 2 x ^{2} - 3 x} - \frac{14 x}{x ^{3} - 2 x ^{2} - 3 x} - \frac{6}{x ^{3} - 2 x ^{2} - 3 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{4 x ^{2}}{x ^{3} - 2 x ^{2} - 3 x} d x - \int \frac{14 x}{x ^{3} - 2 x ^{2} - 3 x} d x - \int \frac{6}{x ^{3} - 2 x ^{2} - 3 x} d x

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

\int \frac{14 x}{x ^{3} - 2 x ^{2} - 3 x} d x = 14 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣)

\int \frac{6}{x ^{3} - 2 x ^{2} - 3 x} d x = 6 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{12} ln ∣ x - 3 ∣)

= ln ∣ x + 1 ∣ + 3 ln ∣ x - 3 ∣ - 14 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣) - 6 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{12} ln ∣ x - 3 ∣)

簡素化

ln ∣ x + 1 ∣ + 3 ln ∣ x - 3 ∣ - 14 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣) - 6 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{12} ln ∣ x - 3 ∣) : 3 ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ + 2 ln ∣ x ∣

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

定数を解答に追加する

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