integral of (2x+1)sqrt(4x-3)

解

\int (2 x + 1) 4 x - 3 d x

\frac{1}{5} x (4 x - 3)^{\frac{3}{2}} + \frac{4}{15} (4 x - 3)^{\frac{3}{2}} + C

解答ステップ

\int (2 x + 1) 4 x - 3 d x

拡張

(2 x + 1) 4 x - 3 : 2 x 4 x - 3 + 4 x - 3

= \int 2 x 4 x - 3 + 4 x - 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 2 x 4 x - 3 d x + \int 4 x - 3 d x

\int 2 x 4 x - 3 d x = \frac{1}{10} (2 x + 1) (4 x - 3)^{\frac{3}{2}}

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

= \frac{1}{10} (2 x + 1) (4 x - 3)^{\frac{3}{2}} + \frac{1}{6} (4 x - 3)^{\frac{3}{2}}

簡素化

\frac{1}{10} (2 x + 1) (4 x - 3)^{\frac{3}{2}} + \frac{1}{6} (4 x - 3)^{\frac{3}{2}} : \frac{1}{5} x (4 x - 3)^{\frac{3}{2}} + \frac{4}{15} (4 x - 3)^{\frac{3}{2}}

= \frac{1}{5} x (4 x - 3)^{\frac{3}{2}} + \frac{4}{15} (4 x - 3)^{\frac{3}{2}}

定数を解答に追加する

= \frac{1}{5} x (4 x - 3)^{\frac{3}{2}} + \frac{4}{15} (4 x - 3)^{\frac{3}{2}} + C