integral of 3x(x^2+8)^2

解

\int 3 x (x^{2} + 8)^{2} d x

3 (\frac{x ^{6}}{6} + 4 x^{4} + 32 x^{2}) + C

解答ステップ

\int 3 x (x^{2} + 8)^{2} d x

定数を除く:

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

= 3 \cdot \int x (x^{2} + 8)^{2} d x

拡張

x (x^{2} + 8)^{2} : x^{5} + 16 x^{3} + 64 x

= 3 \cdot \int x^{5} + 16 x^{3} + 64 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

= 3 (\int x^{5} d x + \int 16 x^{3} d x + \int 64 x d x)

\int x^{5} d x = \frac{x ^{6}}{6}

\int 16 x^{3} d x = 4 x^{4}

\int 64 x d x = 32 x^{2}

= 3 (\frac{x ^{6}}{6} + 4 x^{4} + 32 x^{2})

定数を解答に追加する

= 3 (\frac{x ^{6}}{6} + 4 x^{4} + 32 x^{2}) + C