integral of (x^2-3x+1)/(sqrt(x))

解

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

\frac{2}{5} x^{\frac{5}{2}} - 2 x^{\frac{3}{2}} + 2 x + C

解答ステップ

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

拡張

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

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

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

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

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

= \frac{2}{5} x^{\frac{5}{2}} - 2 x^{\frac{3}{2}} + 2 x

定数を解答に追加する

= \frac{2}{5} x^{\frac{5}{2}} - 2 x^{\frac{3}{2}} + 2 x + C