integral of (1/2 x+3)^2

解

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

\frac{x ^{3}}{12} + \frac{3 x ^{2}}{2} + 9 x + C

解答ステップ

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

簡素化

(\frac{1}{2} x + 3)^{2} : \frac{1}{4} x^{2} + 3 x + 9

= \int \frac{1}{4} x^{2} + 3 x + 9 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 \frac{1}{4} x^{2} d x + \int 3 x d x + \int 9 d x

\int \frac{1}{4} x^{2} d x = \frac{x ^{3}}{12}

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

\int 9 d x = 9 x

= \frac{x ^{3}}{12} + \frac{3 x ^{2}}{2} + 9 x

定数を解答に追加する

= \frac{x ^{3}}{12} + \frac{3 x ^{2}}{2} + 9 x + C