integral of x(sqrt(x)+1)

解

\int x (x + 1) d x

\frac{2}{5} x^{\frac{5}{2}} + \frac{x ^{2}}{2} + C

解答ステップ

\int x (x + 1) d x

拡張

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

= \int x^{\frac{3}{2}} + x 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 x^{\frac{3}{2}} d x + \int x d x

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

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

= \frac{2}{5} x^{\frac{5}{2}} + \frac{x ^{2}}{2}

定数を解答に追加する

= \frac{2}{5} x^{\frac{5}{2}} + \frac{x ^{2}}{2} + C