integral of (6x^2-20x-5)/(x^3+2x^2+x)

解

\int \frac{6 x ^{2} - 20 x - 5}{x ^{3} + 2 x ^{2} + x} d x

11 ln ∣ x + 1 ∣ + \frac{21}{x + 1} - 5 ln ∣ x ∣ + C

解答ステップ

\int \frac{6 x ^{2} - 20 x - 5}{x ^{3} + 2 x ^{2} + x} d x

拡張

\frac{6 x ^{2} - 20 x - 5}{x ^{3} + 2 x ^{2} + x} : \frac{6 x ^{2}}{x ^{3} + 2 x ^{2} + x} - \frac{20 x}{x ^{3} + 2 x ^{2} + x} - \frac{5}{x ^{3} + 2 x ^{2} + 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{6 x ^{2}}{x ^{3} + 2 x ^{2} + x} d x - \int \frac{20 x}{x ^{3} + 2 x ^{2} + x} d x - \int \frac{5}{x ^{3} + 2 x ^{2} + x} d x

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

\int \frac{20 x}{x ^{3} + 2 x ^{2} + x} d x = - \frac{20}{x + 1}

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

= 6 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - (- \frac{20}{x + 1}) - 5 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1})

簡素化

6 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - (- \frac{20}{x + 1}) - 5 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1}) : 11 ln ∣ x + 1 ∣ + \frac{21}{x + 1} - 5 ln ∣ x ∣

= 11 ln ∣ x + 1 ∣ + \frac{21}{x + 1} - 5 ln ∣ x ∣

定数を解答に追加する

= 11 ln ∣ x + 1 ∣ + \frac{21}{x + 1} - 5 ln ∣ x ∣ + C