integral of (3x-1)(sqrt(x)+1)

解

\int (3 x - 1) (x + 1) d x

\frac{6}{5} x^{\frac{5}{2}} + \frac{3 x ^{2}}{2} - \frac{2}{3} x^{\frac{3}{2}} - x + C

解答ステップ

\int (3 x - 1) (x + 1) d x

拡張

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

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

\int 3 x^{\frac{3}{2}} d x = \frac{6}{5} x^{\frac{5}{2}}

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

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

\int 1 d x = x

= \frac{6}{5} x^{\frac{5}{2}} + \frac{3 x ^{2}}{2} - \frac{2}{3} x^{\frac{3}{2}} - x

定数を解答に追加する

= \frac{6}{5} x^{\frac{5}{2}} + \frac{3 x ^{2}}{2} - \frac{2}{3} x^{\frac{3}{2}} - x + C