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

解

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

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

解答ステップ

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

拡張

(x^{3} + 1)^{2} x : x^{7} + 2 x^{4} + x

= 3 \cdot \int x^{7} + 2 x^{4} + 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^{7} d x + \int 2 x^{4} d x + \int x d x)

\int x^{7} d x = \frac{x ^{8}}{8}

\int 2 x^{4} d x = \frac{2 x ^{5}}{5}

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

= 3 (\frac{x ^{8}}{8} + \frac{2 x ^{5}}{5} + \frac{x ^{2}}{2})

定数を解答に追加する

= 3 (\frac{x ^{8}}{8} + \frac{2 x ^{5}}{5} + \frac{x ^{2}}{2}) + C