integral of ((2sqrt(x)-5)^2)/3

解

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

\frac{1}{3} (2 x^{2} - \frac{40}{3} x^{\frac{3}{2}} + 25 x) + C

解答ステップ

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int (2 x - 5)^{2} d x

2 x - 5 = 5 (\frac{2 x}{5} - 1)

= \frac{1}{3} \cdot \int (5 (\frac{2 x}{5} - 1))^{2} d x

簡素化

(5 (\frac{2 x}{5} - 1))^{2} : 4 x - 20 x + 25

= \frac{1}{3} \cdot \int 4 x - 20 x + 25 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

= \frac{1}{3} (\int 4 x d x - \int 20 x d x + \int 25 d x)

\int 4 x d x = 2 x^{2}

\int 20 x d x = \frac{40}{3} x^{\frac{3}{2}}

\int 25 d x = 25 x

= \frac{1}{3} (2 x^{2} - \frac{40}{3} x^{\frac{3}{2}} + 25 x)

定数を解答に追加する

= \frac{1}{3} (2 x^{2} - \frac{40}{3} x^{\frac{3}{2}} + 25 x) + C