integral of (4x+1)sqrt(x-5)

解

\int (4 x + 1) x - 5 d x

\frac{8}{5} (x - 5)^{\frac{5}{2}} + 14 (x - 5)^{\frac{3}{2}} + C

解答ステップ

\int (4 x + 1) x - 5 d x

拡張

(4 x + 1) x - 5 : 4 x x - 5 + x - 5

= \int 4 x x - 5 + x - 5 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 4 x x - 5 d x + \int x - 5 d x

\int 4 x x - 5 d x = 4 (\frac{2}{5} (x - 5)^{\frac{5}{2}} + \frac{10}{3} (x - 5)^{\frac{3}{2}})

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

= 4 (\frac{2}{5} (x - 5)^{\frac{5}{2}} + \frac{10}{3} (x - 5)^{\frac{3}{2}}) + \frac{2}{3} (x - 5)^{\frac{3}{2}}

簡素化

4 (\frac{2}{5} (x - 5)^{\frac{5}{2}} + \frac{10}{3} (x - 5)^{\frac{3}{2}}) + \frac{2}{3} (x - 5)^{\frac{3}{2}} : \frac{8}{5} (x - 5)^{\frac{5}{2}} + 14 (x - 5)^{\frac{3}{2}}

= \frac{8}{5} (x - 5)^{\frac{5}{2}} + 14 (x - 5)^{\frac{3}{2}}

定数を解答に追加する

= \frac{8}{5} (x - 5)^{\frac{5}{2}} + 14 (x - 5)^{\frac{3}{2}} + C