integral of (sqrt(x)+1/(\sqrt[3]{x)})^3

解

\int (x + \frac{1}{3 x})^{3} d x

ln ∣ x ∣ + \frac{18}{5} x^{\frac{5}{6}} + \frac{9}{5} x^{\frac{5}{3}} + \frac{2}{5} x^{\frac{5}{2}} + C

解答ステップ

\int (x + \frac{1}{3 x})^{3} d x

拡張

(x + \frac{1}{3 x})^{3} : \frac{1}{x} + \frac{3}{x ^{\frac{1}{6}}} + 3 x^{\frac{2}{3}} + x^{\frac{3}{2}}

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{3}{x ^{\frac{1}{6}}} d x = \frac{18}{5} x^{\frac{5}{6}}

\int 3 x^{\frac{2}{3}} d x = \frac{9}{5} x^{\frac{5}{3}}

\int x^{\frac{3}{2}} d x = \frac{2}{5} x^{\frac{5}{2}}

= ln ∣ x ∣ + \frac{18}{5} x^{\frac{5}{6}} + \frac{9}{5} x^{\frac{5}{3}} + \frac{2}{5} x^{\frac{5}{2}}

定数を解答に追加する

= ln ∣ x ∣ + \frac{18}{5} x^{\frac{5}{6}} + \frac{9}{5} x^{\frac{5}{3}} + \frac{2}{5} x^{\frac{5}{2}} + C