integral of (10x+1)sqrt(4x-1)

解

\int (10 x + 1) 4 x - 1 d x

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

解答ステップ

\int (10 x + 1) 4 x - 1 d x

拡張

(10 x + 1) 4 x - 1 : 10 x 4 x - 1 + 4 x - 1

= \int 10 x 4 x - 1 + 4 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 10 x 4 x - 1 d x + \int 4 x - 1 d x

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

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

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

簡素化

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

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

定数を解答に追加する

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