integral of 5x(x^2+4)^2

解

\int 5 x (x^{2} + 4)^{2} d x

5 (\frac{x ^{6}}{6} + 2 x^{4} + 8 x^{2}) + C

解答ステップ

\int 5 x (x^{2} + 4)^{2} d x

定数を除く:

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

= 5 \cdot \int x (x^{2} + 4)^{2} d x

拡張

x (x^{2} + 4)^{2} : x^{5} + 8 x^{3} + 16 x

= 5 \cdot \int x^{5} + 8 x^{3} + 16 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

= 5 (\int x^{5} d x + \int 8 x^{3} d x + \int 16 x d x)

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

\int 8 x^{3} d x = 2 x^{4}

\int 16 x d x = 8 x^{2}

= 5 (\frac{x ^{6}}{6} + 2 x^{4} + 8 x^{2})

定数を解答に追加する

= 5 (\frac{x ^{6}}{6} + 2 x^{4} + 8 x^{2}) + C