integral of (x^2-3x+\sqrt[3]{x})/(x^3)

解

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

ln ∣ x ∣ + \frac{3}{x} - \frac{3}{5 x ^{\frac{5}{3}}} + C

解答ステップ

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

拡張

\frac{x ^{2} - 3 x + 3 x}{x ^{3}} : \frac{1}{x} - \frac{3}{x ^{2}} + \frac{1}{x ^{\frac{8}{3}}}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int \frac{1}{x ^{\frac{8}{3}}} d x = - \frac{3}{5 x ^{\frac{5}{3}}}

= ln ∣ x ∣ - (- \frac{3}{x}) - \frac{3}{5 x ^{\frac{5}{3}}}

簡素化

= ln ∣ x ∣ + \frac{3}{x} - \frac{3}{5 x ^{\frac{5}{3}}}

定数を解答に追加する

= ln ∣ x ∣ + \frac{3}{x} - \frac{3}{5 x ^{\frac{5}{3}}} + C