integral of ((3x+1)^2)/(244)

解

\int \frac{( 3 x + 1 ) ^{2}}{244} d x

\frac{1}{244} (3 x^{3} + 3 x^{2} + x) + C

解答ステップ

\int \frac{( 3 x + 1 ) ^{2}}{244} d x

定数を除く:

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

= \frac{1}{244} \cdot \int (3 x + 1)^{2} d x

拡張

(3 x + 1)^{2} : 9 x^{2} + 6 x + 1

= \frac{1}{244} \cdot \int 9 x^{2} + 6 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

= \frac{1}{244} (\int 9 x^{2} d x + \int 6 x d x + \int 1 d x)

\int 9 x^{2} d x = 3 x^{3}

\int 6 x d x = 3 x^{2}

\int 1 d x = x

= \frac{1}{244} (3 x^{3} + 3 x^{2} + x)

定数を解答に追加する

= \frac{1}{244} (3 x^{3} + 3 x^{2} + x) + C