integral of (x^4+3x+1)/(x^3+x^2-2x)

解

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

\frac{x ^{2}}{2} - x + \frac{2}{3} ln ∣ x - 1 ∣ + \frac{17}{6} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x ∣ - ln \frac{2}{3} x + \frac{4}{3} + ln \frac{2}{3} x - \frac{2}{3} + C

解答ステップ

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

拡張

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

\int \frac{x ^{4}}{x ^{3} + x ^{2} - 2 x} d x = \frac{x ^{2}}{2} - x + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{8}{3} ln ∣ x + 2 ∣

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

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

= \frac{x ^{2}}{2} - x + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{8}{3} ln ∣ x + 2 ∣ - ln \frac{4}{3} + \frac{2}{3} x + ln - \frac{2}{3} + \frac{2}{3} x - \frac{1}{2} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x + 2 ∣

簡素化

\frac{x ^{2}}{2} - x + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{8}{3} ln ∣ x + 2 ∣ - ln \frac{4}{3} + \frac{2}{3} x + ln - \frac{2}{3} + \frac{2}{3} x - \frac{1}{2} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x + 2 ∣ : \frac{x ^{2}}{2} - x + \frac{2}{3} ln ∣ x - 1 ∣ + \frac{17}{6} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x ∣ - ln \frac{2}{3} x + \frac{4}{3} + ln \frac{2}{3} x - \frac{2}{3}

= \frac{x ^{2}}{2} - x + \frac{2}{3} ln ∣ x - 1 ∣ + \frac{17}{6} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x ∣ - ln \frac{2}{3} x + \frac{4}{3} + ln \frac{2}{3} x - \frac{2}{3}

定数を解答に追加する

= \frac{x ^{2}}{2} - x + \frac{2}{3} ln ∣ x - 1 ∣ + \frac{17}{6} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x ∣ - ln \frac{2}{3} x + \frac{4}{3} + ln \frac{2}{3} x - \frac{2}{3} + C