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

解

\int (2 x + 3) 2 x - 1 d x

\frac{2}{5} x (2 x - 1)^{\frac{3}{2}} + \frac{17}{15} (2 x - 1)^{\frac{3}{2}} + C

解答ステップ

\int (2 x + 3) 2 x - 1 d x

拡張

(2 x + 3) 2 x - 1 : 2 x 2 x - 1 + 3 2 x - 1

= \int 2 x 2 x - 1 + 3 2 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 2 x 2 x - 1 d x + \int 3 2 x - 1 d x

\int 2 x 2 x - 1 d x = \frac{2}{15} (3 x + 1) (2 x - 1)^{\frac{3}{2}}

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

= \frac{2}{15} (3 x + 1) (2 x - 1)^{\frac{3}{2}} + (2 x - 1)^{\frac{3}{2}}

簡素化

\frac{2}{15} (3 x + 1) (2 x - 1)^{\frac{3}{2}} + (2 x - 1)^{\frac{3}{2}} : \frac{2}{5} x (2 x - 1)^{\frac{3}{2}} + \frac{17}{15} (2 x - 1)^{\frac{3}{2}}

= \frac{2}{5} x (2 x - 1)^{\frac{3}{2}} + \frac{17}{15} (2 x - 1)^{\frac{3}{2}}

定数を解答に追加する

= \frac{2}{5} x (2 x - 1)^{\frac{3}{2}} + \frac{17}{15} (2 x - 1)^{\frac{3}{2}} + C