integral of (4x^2-7x-12)/(x^3+5x^2+6x)

解

\int \frac{4 x ^{2} - 7 x - 12}{x ^{3} + 5 x ^{2} + 6 x} d x

- 9 ln ∣ x + 2 ∣ + 15 ln ∣ x + 3 ∣ - 2 ln ∣ x ∣ + C

解答ステップ

\int \frac{4 x ^{2} - 7 x - 12}{x ^{3} + 5 x ^{2} + 6 x} d x

拡張

\frac{4 x ^{2} - 7 x - 12}{x ^{3} + 5 x ^{2} + 6 x} : \frac{4 x ^{2}}{x ^{3} + 5 x ^{2} + 6 x} - \frac{7 x}{x ^{3} + 5 x ^{2} + 6 x} - \frac{12}{x ^{3} + 5 x ^{2} + 6 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} + 5 x ^{2} + 6 x} d x - \int \frac{7 x}{x ^{3} + 5 x ^{2} + 6 x} d x - \int \frac{12}{x ^{3} + 5 x ^{2} + 6 x} d x

\int \frac{4 x ^{2}}{x ^{3} + 5 x ^{2} + 6 x} d x = 4 (- 2 ln ∣ x + 2 ∣ + 3 ln ∣ x + 3 ∣)

\int \frac{7 x}{x ^{3} + 5 x ^{2} + 6 x} d x = 7 (ln ∣ x + 2 ∣ - ln ∣ x + 3 ∣)

\int \frac{12}{x ^{3} + 5 x ^{2} + 6 x} d x = 12 (\frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x + 3 ∣)

= 4 (- 2 ln ∣ x + 2 ∣ + 3 ln ∣ x + 3 ∣) - 7 (ln ∣ x + 2 ∣ - ln ∣ x + 3 ∣) - 12 (\frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x + 3 ∣)

簡素化

4 (- 2 ln ∣ x + 2 ∣ + 3 ln ∣ x + 3 ∣) - 7 (ln ∣ x + 2 ∣ - ln ∣ x + 3 ∣) - 12 (\frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 2 ∣ + \frac{1}{3} ln ∣ x + 3 ∣) : - 9 ln ∣ x + 2 ∣ + 15 ln ∣ x + 3 ∣ - 2 ln ∣ x ∣

= - 9 ln ∣ x + 2 ∣ + 15 ln ∣ x + 3 ∣ - 2 ln ∣ x ∣

定数を解答に追加する

= - 9 ln ∣ x + 2 ∣ + 15 ln ∣ x + 3 ∣ - 2 ln ∣ x ∣ + C