integral of (x^2+1/(\sqrt[3]{x)})^2

解

\int (x^{2} + \frac{1}{3 x})^{2} d x

3 3 x + \frac{3}{4} x^{\frac{8}{3}} + \frac{x ^{5}}{5} + C

解答ステップ

\int (x^{2} + \frac{1}{3 x})^{2} d x

拡張

(x^{2} + \frac{1}{3 x})^{2} : \frac{1}{x ^{\frac{2}{3}}} + 2 x^{\frac{5}{3}} + x^{4}

= \int \frac{1}{x ^{\frac{2}{3}}} + 2 x^{\frac{5}{3}} + x^{4} 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 ^{\frac{2}{3}}} d x + \int 2 x^{\frac{5}{3}} d x + \int x^{4} d x

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

\int 2 x^{\frac{5}{3}} d x = \frac{3}{4} x^{\frac{8}{3}}

\int x^{4} d x = \frac{x ^{5}}{5}

= 3 x^{\frac{1}{3}} + \frac{3}{4} x^{\frac{8}{3}} + \frac{x ^{5}}{5}

累乗根の規則を適用する:

a^{\frac{1}{n}} = n a

= 3 3 x + \frac{3}{4} x^{\frac{8}{3}} + \frac{x ^{5}}{5}

定数を解答に追加する

= 3 3 x + \frac{3}{4} x^{\frac{8}{3}} + \frac{x ^{5}}{5} + C