integral of (x^4)/(x^3-8)

解

\int \frac{x ^{4}}{x ^{3} - 8} d x

\frac{x ^{2}}{2} + 8 (\frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{12} ln x^{2} + 2 x + 4 + \frac{1}{2 3} arctan (\frac{1}{3} (x + 1))) + C

解答ステップ

\int \frac{x ^{4}}{x ^{3} - 8} d x

長除法

\frac{x ^{4}}{x ^{3} - 8} : x + \frac{8 x}{x ^{3} - 8}

= \int x + \frac{8 x}{x ^{3} - 8} d 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 x d x + \int \frac{8 x}{x ^{3} - 8} d x

\int x d x = \frac{x ^{2}}{2}

\int \frac{8 x}{x ^{3} - 8} d x = 8 (\frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{12} ln x^{2} + 2 x + 4 + \frac{1}{2 3} arctan (\frac{1}{3} (x + 1)))

= \frac{x ^{2}}{2} + 8 (\frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{12} ln x^{2} + 2 x + 4 + \frac{1}{2 3} arctan (\frac{1}{3} (x + 1)))

定数を解答に追加する

= \frac{x ^{2}}{2} + 8 (\frac{1}{6} ln ∣ x - 2 ∣ - \frac{1}{12} ln x^{2} + 2 x + 4 + \frac{1}{2 3} arctan (\frac{1}{3} (x + 1))) + C