integral of (x^3-6x+18)/(3x^{1/2)}

解

\int \frac{x ^{3} - 6 x + 18}{3 x ^{\frac{1}{2}}} d x

\frac{1}{3} (\frac{2}{7} x^{\frac{7}{2}} - 4 x^{\frac{3}{2}} + 36 x^{\frac{1}{2}}) + C

解答ステップ

\int \frac{x ^{3} - 6 x + 18}{3 x ^{\frac{1}{2}}} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{x ^{3} - 6 x + 18}{x ^{\frac{1}{2}}} d x

拡張

\frac{x ^{3} - 6 x + 18}{x ^{\frac{1}{2}}} : x^{\frac{5}{2}} - 6 x^{\frac{1}{2}} + \frac{18}{x ^{\frac{1}{2}}}

= \frac{1}{3} \cdot \int x^{\frac{5}{2}} - 6 x^{\frac{1}{2}} + \frac{18}{x ^{\frac{1}{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

= \frac{1}{3} (\int x^{\frac{5}{2}} d x - \int 6 x^{\frac{1}{2}} d x + \int \frac{18}{x ^{\frac{1}{2}}} d x)

\int x^{\frac{5}{2}} d x = \frac{2}{7} x^{\frac{7}{2}}

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

\int \frac{18}{x ^{\frac{1}{2}}} d x = 36 x^{\frac{1}{2}}

= \frac{1}{3} (\frac{2}{7} x^{\frac{7}{2}} - 4 x^{\frac{3}{2}} + 36 x^{\frac{1}{2}})

定数を解答に追加する

= \frac{1}{3} (\frac{2}{7} x^{\frac{7}{2}} - 4 x^{\frac{3}{2}} + 36 x^{\frac{1}{2}}) + C