integral of (sqrt(n)+3)^3

解

\int (n + 3)^{3} d n

\frac{2}{5} n^{\frac{5}{2}} + \frac{9 n ^{2}}{2} + 18 n^{\frac{3}{2}} + 27 n + C

解答ステップ

\int (n + 3)^{3} d n

拡張

(n + 3)^{3} : n^{\frac{3}{2}} + 9 n + 27 n + 27

= \int n^{\frac{3}{2}} + 9 n + 27 n + 27 d n

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 n^{\frac{3}{2}} d n + \int 9 n d n + \int 27 n d n + \int 27 d n

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

\int 9 n d n = \frac{9 n ^{2}}{2}

\int 27 n d n = 18 n^{\frac{3}{2}}

\int 27 d n = 27 n

= \frac{2}{5} n^{\frac{5}{2}} + \frac{9 n ^{2}}{2} + 18 n^{\frac{3}{2}} + 27 n

定数を解答に追加する

= \frac{2}{5} n^{\frac{5}{2}} + \frac{9 n ^{2}}{2} + 18 n^{\frac{3}{2}} + 27 n + C