integral of (x^2+1)nx

解

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

n (\frac{x ^{4}}{4} + \frac{x ^{2}}{2}) + C

解答ステップ

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

定数を除く:

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

= n \cdot \int (x^{2} + 1) x d x

拡張

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

= n \cdot \int x^{3} + 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

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

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

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

= n (\frac{x ^{4}}{4} + \frac{x ^{2}}{2})

定数を解答に追加する

= n (\frac{x ^{4}}{4} + \frac{x ^{2}}{2}) + C