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

解

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

ln ∣ x ∣ + \frac{12}{6 x} - \frac{3}{3 x} + C

解答ステップ

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

拡張

(\frac{x - 3 x}{x})^{2} : \frac{1}{x} - \frac{2}{x ^{\frac{7}{6}}} + \frac{1}{x ^{\frac{4}{3}}}

= \int \frac{1}{x} - \frac{2}{x ^{\frac{7}{6}}} + \frac{1}{x ^{\frac{4}{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{2}{x ^{\frac{7}{6}}} d x + \int \frac{1}{x ^{\frac{4}{3}}} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{2}{x ^{\frac{7}{6}}} d x = - \frac{12}{6 x}

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

= ln ∣ x ∣ - (- \frac{12}{6 x}) - \frac{3}{3 x}

簡素化

= ln ∣ x ∣ + \frac{12}{6 x} - \frac{3}{3 x}

定数を解答に追加する

= ln ∣ x ∣ + \frac{12}{6 x} - \frac{3}{3 x} + C